Chapter 4 Quadratic Equations

Solution:

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A quadratic equation is an equation that can be written in the standard form $ax^2 + bx + c = 0$, where $x$ is the variable, and $a$, $b$, $c$ are coefficients, with $a \neq 0$. We will examine each option by simplifying it to see if it fits this form.

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(A) $(x + 2)^2 = 2(x + 3)$

Expanding the left side using $(a+b)^2 = a^2 + 2ab + b^2$:

$(x + 2)^2 = x^2 + 2(x)(2) + 2^2 = x^2 + 4x + 4$

Expanding the right side:

$2(x + 3) = 2x + 6$

Equating the expanded forms:

$x^2 + 4x + 4 = 2x + 6$

Rearranging all terms to one side:

$x^2 + 4x - 2x + 4 - 6 = 0$

$x^2 + 2x - 2 = 0$

This equation is in the form $ax^2 + bx + c = 0$ with $a=1$, $b=2$, and $c=-2$. Since $a=1 \neq 0$, this is a quadratic equation.

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(B) $x^2 + 3x = (-1) (1 - 3x)^2$

Expanding the term $(1 - 3x)^2$ using $(a-b)^2 = a^2 - 2ab + b^2$:

$(1 - 3x)^2 = 1^2 - 2(1)(3x) + (3x)^2 = 1 - 6x + 9x^2$

Multiplying by $(-1)$:

$(-1)(1 - 3x)^2 = -(1 - 6x + 9x^2) = -1 + 6x - 9x^2$

Equating the left and right sides:

$x^2 + 3x = -1 + 6x - 9x^2$

Rearranging all terms to one side:

$x^2 + 9x^2 + 3x - 6x + 1 = 0$

$10x^2 - 3x + 1 = 0$

This equation is in the form $ax^2 + bx + c = 0$ with $a=10$, $b=-3$, and $c=1$. Since $a=10 \neq 0$, this is a quadratic equation.

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(C) $(x + 2) (x - 1) = x^2 - 2x - 3$

Expanding the left side:

$(x + 2) (x - 1) = x(x - 1) + 2(x - 1) = x^2 - x + 2x - 2 = x^2 + x - 2$

Equating the left and right sides:

$x^2 + x - 2 = x^2 - 2x - 3$

Rearranging all terms to one side:

$x^2 - x^2 + x + 2x - 2 + 3 = 0$

$0x^2 + 3x + 1 = 0$

$3x + 1 = 0$

In this equation, the coefficient of $x^2$ is $a=0$. Therefore, this equation is of degree 1 and is a linear equation, not a quadratic equation.

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(D) $x^3 - x^2 + 2x + 1 = (x + 1)^3$

Expanding the right side using $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$:

$(x + 1)^3 = x^3 + 3(x^2)(1) + 3(x)(1^2) + 1^3 = x^3 + 3x^2 + 3x + 1$

Equating the left and right sides:

$x^3 - x^2 + 2x + 1 = x^3 + 3x^2 + 3x + 1$

Rearranging all terms to one side:

$x^3 - x^3 - x^2 - 3x^2 + 2x - 3x + 1 - 1 = 0$

$0x^3 - 4x^2 - x + 0 = 0$

$-4x^2 - x = 0$

This equation can be written as $-4x^2 - x + 0 = 0$, which is in the form $ax^2 + bx + c = 0$ with $a=-4$, $b=-1$, and $c=0$. Since $a=-4 \neq 0$, this is a quadratic equation.

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Based on the analysis, only option (C) simplifies to a linear equation ($3x+1=0$). All other options simplify to quadratic equations.

Thus, the equation that is not a quadratic equation is (C) $(x + 2) (x - 1) = x^2 - 2x - 3$.

Solution:

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The given quadratic equation is $4x^2 - \sqrt{3}x - 5 = 0$.

The method of completing the square involves transforming the part of the equation containing the $x^2$ and $x$ terms into a perfect square trinomial. For a general quadratic equation $ax^2 + bx + c = 0$, we focus on the terms $ax^2 + bx$.

We can factor out the coefficient of $x^2$, $a$, from the first two terms:

$ax^2 + bx = a(x^2 + \frac{b}{a}x)$

To complete the square for the expression inside the parenthesis, $x^2 + \frac{b}{a}x$, we need to add the square of half of the coefficient of the $x$ term. The coefficient of $x$ is $\frac{b}{a}$. Half of this coefficient is $\frac{1}{2} \times \frac{b}{a} = \frac{b}{2a}$.

Squaring this value gives $(\frac{b}{2a})^2 = \frac{b^2}{4a^2}$.

Adding this term inside the parenthesis creates a perfect square trinomial:

$x^2 + \frac{b}{a}x + \frac{b^2}{4a^2} = (x + \frac{b}{2a})^2$

Now, substitute this back into the expression $a(x^2 + \frac{b}{a}x)$:

$a(x^2 + \frac{b}{a}x + \frac{b^2}{4a^2}) = a(x + \frac{b}{2a})^2$

Expanding the left side:

$ax^2 + bx + \frac{ab^2}{4a^2} = a(x + \frac{b}{2a})^2$

$ax^2 + bx + \frac{b^2}{4a} = a(x + \frac{b}{2a})^2$

This shows that to make the terms $ax^2 + bx$ part of a perfect square form $a(x + \frac{b}{2a})^2$, the constant $\frac{b^2}{4a}$ must be added.

To solve the equation $ax^2 + bx + c = 0$ by the method of completing the square, we add and subtract this constant $\frac{b^2}{4a}$ to the original equation to maintain equality:

$ax^2 + bx + \frac{b^2}{4a} - \frac{b^2}{4a} + c = 0$

The first three terms form the perfect square: $(ax^2 + bx + \frac{b^2}{4a}) - \frac{b^2}{4a} + c = 0$, which is $a(x + \frac{b}{2a})^2 - \frac{b^2}{4a} + c = 0$.

The constant that is added and subtracted to the original equation is $\frac{b^2}{4a}$.

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Now, we apply this to the given equation $4x^2 - \sqrt{3}x - 5 = 0$.

Comparing this to the standard form $ax^2 + bx + c = 0$, we have $a=4$ and $b=-\sqrt{3}$.

The constant to be added and subtracted is $k = \frac{b^2}{4a}$.

Substituting the values of $a$ and $b$:

$k = \frac{(-\sqrt{3})^2}{4 \times 4}$

$k = \frac{3}{16}$

So, the constant that should be added and subtracted to solve the equation by completing the square is $\frac{3}{16}$.

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Let's compare this result with the given options:

(A) $\frac{9}{16}$

(B) $\frac{3}{16}$

(C) $\frac{3}{4}$

(D) $\frac{\sqrt{3}}{4}$

The calculated constant $\frac{3}{16}$ matches option (B).

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The correct answer is (B) $\frac{3}{16}$.

Solution:

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A quadratic equation is an equation of the form $ax^2 + bx + c = 0$, where $x$ is the variable, and $a$, $b$, and $c$ are real numbers with $a \neq 0$. We need to simplify each given option and check if it fits this form.

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(A) $x^2 + 2x + 1 = (4 – x)^2 + 3$

Expanding the right side: $(4 – x)^2 = 4^2 - 2(4)(x) + x^2 = 16 - 8x + x^2$.

So, the equation becomes: $x^2 + 2x + 1 = 16 - 8x + x^2 + 3$

Simplifying: $x^2 + 2x + 1 = x^2 - 8x + 19$

Moving all terms to one side: $x^2 - x^2 + 2x + 8x + 1 - 19 = 0$

$10x - 18 = 0$

This is a linear equation because the coefficient of $x^2$ is 0. It is not a quadratic equation.

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(B) $-2x^2 = (5 – x)\left( 2x - \frac{2}{5} \right)$

Expanding the right side:

$(5 – x)\left( 2x - \frac{2}{5} \right) = 5(2x) + 5\left(-\frac{2}{5}\right) - x(2x) - x\left(-\frac{2}{5}\right)$

$= 10x - 2 - 2x^2 + \frac{2}{5}x$

Combining like terms on the right side: $-2x^2 + \left(10 + \frac{2}{5}\right)x - 2$

$10 + \frac{2}{5} = \frac{50}{5} + \frac{2}{5} = \frac{52}{5}$

So, the right side is $-2x^2 + \frac{52}{5}x - 2$.

The equation becomes: $-2x^2 = -2x^2 + \frac{52}{5}x - 2$

Moving all terms to one side: $-2x^2 + 2x^2 - \frac{52}{5}x + 2 = 0$

$-\frac{52}{5}x + 2 = 0$

This is a linear equation because the coefficient of $x^2$ is 0. It is not a quadratic equation.

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(C) $(k + 1)x^2 + \frac{3}{2} x = 7$, where $k = –1$

Substitute the value of $k = -1$ into the equation:

$(-1 + 1)x^2 + \frac{3}{2}x = 7$

$(0)x^2 + \frac{3}{2}x = 7$

$\frac{3}{2}x = 7$

This is a linear equation because the coefficient of $x^2$ is 0. It is not a quadratic equation.

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(D) $x^3 – x^2 = (x – 1)^3$

Expanding the right side using the identity $(a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$:

$(x – 1)^3 = x^3 - 3(x^2)(1) + 3(x)(1^2) - 1^3 = x^3 - 3x^2 + 3x - 1$

The equation becomes: $x^3 - x^2 = x^3 - 3x^2 + 3x - 1$

Moving all terms to one side: $x^3 - x^3 - x^2 + 3x^2 - 3x + 1 = 0$

Combining like terms: $2x^2 - 3x + 1 = 0$

This equation is in the form $ax^2 + bx + c = 0$ with $a=2$, $b=-3$, and $c=1$. Since $a=2 \neq 0$, this is a quadratic equation.

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Based on the analysis, only option (D) results in an equation of the form $ax^2 + bx + c = 0$ with $a \neq 0$.

The correct answer is (D) x³ – x² = (x – 1)³.

Solution:

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A quadratic equation is an equation of the form $ax^2 + bx + c = 0$, where $x$ is the variable, and $a$, $b$, and $c$ are real numbers with $a \neq 0$. We will examine each option to determine if it is a quadratic equation.

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(A) $2(x – 1)^2 = 4x^2 – 2x + 1$

Expanding the left side: $2(x^2 - 2x + 1) = 2x^2 - 4x + 2$

The equation becomes: $2x^2 - 4x + 2 = 4x^2 - 2x + 1$

Moving all terms to the left side:

$2x^2 - 4x^2 - 4x + 2x + 2 - 1 = 0$

$-2x^2 - 2x + 1 = 0$

This equation is of the form $ax^2 + bx + c = 0$ with $a=-2$, $b=-2$, and $c=1$. Since $a=-2 \neq 0$, this is a quadratic equation.

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(B) $2x – x^2 = x^2 + 5$

Moving all terms to the left side:

$2x - x^2 - x^2 - 5 = 0$

$-2x^2 + 2x - 5 = 0$

This equation is of the form $ax^2 + bx + c = 0$ with $a=-2$, $b=2$, and $c=-5$. Since $a=-2 \neq 0$, this is a quadratic equation.

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(C) $(\sqrt{2}x + \sqrt{3})^2 + x^2 = 3x^2 - 5x$

Expanding the term $(\sqrt{2}x + \sqrt{3})^2$ using $(a+b)^2 = a^2 + 2ab + b^2$:

$(\sqrt{2}x)^2 + 2(\sqrt{2}x)(\sqrt{3}) + (\sqrt{3})^2 = 2x^2 + 2\sqrt{6}x + 3$

The equation becomes: $2x^2 + 2\sqrt{6}x + 3 + x^2 = 3x^2 - 5x$

Combining like terms on the left side:

$3x^2 + 2\sqrt{6}x + 3 = 3x^2 - 5x$

Moving all terms to the left side:

$3x^2 - 3x^2 + 2\sqrt{6}x + 5x + 3 = 0$

$(2\sqrt{6} + 5)x + 3 = 0$

In this equation, the coefficient of the $x^2$ term is 0. This is a linear equation, not a quadratic equation.

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(D) $(x^2 + 2x)^2 = x^4 + 3 + 4x^3$

Expanding the left side using $(a+b)^2 = a^2 + 2ab + b^2$:

$(x^2)^2 + 2(x^2)(2x) + (2x)^2 = x^4 + 4x^3 + 4x^2$

The equation becomes: $x^4 + 4x^3 + 4x^2 = x^4 + 3 + 4x^3$

Moving all terms to the left side:

$x^4 - x^4 + 4x^3 - 4x^3 + 4x^2 - 3 = 0$

$4x^2 - 3 = 0$

This equation can be written as $4x^2 + 0x - 3 = 0$, which is of the form $ax^2 + bx + c = 0$ with $a=4$, $b=0$, and $c=-3$. Since $a=4 \neq 0$, this is a quadratic equation.

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Based on the simplification, option (C) is the only equation that is not a quadratic equation as the coefficient of $x^2$ becomes 0.

The correct answer is (C) ($\sqrt{2}$x + $\sqrt{3}$)² + x² = 3x² - 5x .

Solution:

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A number is a root of a quadratic equation if substituting the number for the variable makes the equation true (i.e., satisfies the equation).

We are given four quadratic equations and asked to find which one has $x=2$ as a root. We will substitute $x=2$ into each equation and check if the left side equals the right side (which is 0 in all these cases).

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(A) $x^2 – 4x + 5 = 0$

Substitute $x=2$:

$(2)^2 - 4(2) + 5$

$= 4 - 8 + 5$

$= -4 + 5$

$= 1$

Since $1 \neq 0$, $x=2$ is not a root of this equation.

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(B) $x^2 + 3x – 12 = 0$

Substitute $x=2$:

$(2)^2 + 3(2) - 12$

$= 4 + 6 - 12$

$= 10 - 12$

$= -2$

Since $-2 \neq 0$, $x=2$ is not a root of this equation.

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(C) $2x^2 – 7x + 6 = 0$

Substitute $x=2$:

$2(2)^2 - 7(2) + 6$

$= 2(4) - 14 + 6$

$= 8 - 14 + 6$

$= -6 + 6$

$= 0$

Since $0 = 0$, $x=2$ is a root of this equation.

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(D) $3x^2 – 6x – 2 = 0$

Substitute $x=2$:

$3(2)^2 - 6(2) - 2$

$= 3(4) - 12 - 2$

$= 12 - 12 - 2$

$= 0 - 2$

$= -2$

Since $-2 \neq 0$, $x=2$ is not a root of this equation.

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Therefore, the equation which has 2 as a root is $2x^2 – 7x + 6 = 0$.

The correct answer is (C) 2x² – 7x + 6 = 0.

Given:

The quadratic equation is $x^2 + kx - \frac{5}{4} = 0$.

One of the roots of the equation is $x = \frac{1}{2}$.

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To Find:

The value of the constant $k$.

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Solution:

If $x = \frac{1}{2}$ is a root of the equation $x^2 + kx - \frac{5}{4} = 0$, then substituting $x = \frac{1}{2}$ into the equation must satisfy the equation.

Substitute $x = \frac{1}{2}$ into the equation:

$(\frac{1}{2})^2 + k(\frac{1}{2}) - \frac{5}{4} = 0$

Calculate the square of $\frac{1}{2}$:

$\frac{1}{4} + \frac{k}{2} - \frac{5}{4} = 0$

Combine the constant terms $\frac{1}{4}$ and $-\frac{5}{4}$:

$\frac{1 - 5}{4} + \frac{k}{2} = 0$

$\frac{-4}{4} + \frac{k}{2} = 0$

$-1 + \frac{k}{2} = 0$

Add 1 to both sides of the equation:

$\frac{k}{2} = 1$

Multiply both sides by 2 to solve for $k$:

$k = 1 \times 2$

$k = 2$

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Thus, the value of $k$ is 2.

Comparing this value with the given options:

(A) 2

(B) – 2

(C) $\frac{1}{4}$

(D) $\frac{1}{2}$

The calculated value $k=2$ matches option (A).

The correct answer is (A) 2.

Solution:

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For a standard quadratic equation in the form $ax^2 + bx + c = 0$, where $a \neq 0$, the sum of its roots ($\alpha + \beta$) is given by the formula: $\alpha + \beta = -\frac{b}{a}$.

We will calculate the sum of roots for each given equation and check which one equals 3.

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(A) $2x^2 – 3x + 6 = 0$

Here, $a = 2$, $b = -3$, $c = 6$.

Sum of roots $= -\frac{b}{a} = - \frac{-3}{2} = \frac{3}{2}$.

Since $\frac{3}{2} \neq 3$, this equation does not have the sum of its roots as 3.

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(B) $-x^2 + 3x – 3 = 0$

This equation can be written as $-1x^2 + 3x - 3 = 0$.

Here, $a = -1$, $b = 3$, $c = -3$.

Sum of roots $= -\frac{b}{a} = - \frac{3}{-1} = -(-3) = 3$.

The sum of roots is 3. This equation satisfies the condition.

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(C) $\sqrt{2}x^2 - \frac{3}{\sqrt{2}} x + 1 = 0$

Here, $a = \sqrt{2}$, $b = -\frac{3}{\sqrt{2}}$, $c = 1$.

Sum of roots $= -\frac{b}{a} = - \frac{-\frac{3}{\sqrt{2}}}{\sqrt{2}}$.

Sum of roots $= - \left( -\frac{3}{\sqrt{2}} \times \frac{1}{\sqrt{2}} \right) = - \left( -\frac{3}{(\sqrt{2})^2} \right) = - \left( -\frac{3}{2} \right) = \frac{3}{2}$.

Since $\frac{3}{2} \neq 3$, this equation does not have the sum of its roots as 3.

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(D) $3x^2 – 3x + 3 = 0$

Here, $a = 3$, $b = -3$, $c = 3$.

Sum of roots $= -\frac{b}{a} = - \frac{-3}{3} = -(-1) = 1$.

Since $1 \neq 3$, this equation does not have the sum of its roots as 3.

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Only option (B) has the sum of its roots equal to 3.

The correct answer is (B) –x² + 3x – 3 = 0.

Given:

The quadratic equation is $2x^2 - kx + k = 0$.

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To Find:

The values of $k$ for which the equation has equal roots.

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Solution:

For a quadratic equation in the standard form $ax^2 + bx + c = 0$, the nature of the roots is determined by the discriminant, $\Delta = b^2 - 4aC$.

For the given equation $2x^2 - kx + k = 0$, we compare it to the standard form and identify the coefficients:

$a = 2$

$b = -k$

$c = k$

The condition for a quadratic equation to have equal roots is that the discriminant must be equal to zero ($\Delta = 0$).

So, we set the discriminant to 0:

$b^2 - 4ac = 0$

Substitute the values of $a$, $b$, and $c$ into the discriminant equation:

$(-k)^2 - 4(2)(k) = 0$

Simplify the equation:

$k^2 - 8k = 0$

Factor out $k$ from the equation:

$k(k - 8) = 0$

For the product of two factors to be zero, at least one of the factors must be zero.

So, either $k = 0$ or $k - 8 = 0$.

If $k - 8 = 0$, then $k = 8$.

Thus, the values of $k$ for which the quadratic equation has equal roots are $k = 0$ and $k = 8$.

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Comparing these values with the given options:

(A) 0 only

(B) 4

(C) 8 only

(D) 0, 8

The calculated values $k=0$ and $k=8$ match option (D).

The correct answer is (D) 0, 8.

Solution:

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The given quadratic equation is $9x^2 + \frac{3}{4}x - \sqrt{2} = 0$.

To solve a quadratic equation $ax^2 + bx + c = 0$ by the method of completing the square, we need to add and subtract a specific constant to the terms involving $x^2$ and $x$ to form a perfect square trinomial.

The method involves factoring out the coefficient of $x^2$ (which is $a$) from the first two terms and then adding the square of half the coefficient of $x$ inside the parenthesis. The constant to be added and subtracted to the original equation is $\frac{b^2}{4a}$.

In the given equation, $9x^2 + \frac{3}{4}x - \sqrt{2} = 0$, we have:

$a = 9$

$b = \frac{3}{4}$

$c = -\sqrt{2}$

The constant to be added and subtracted is $\frac{b^2}{4a}$.

Substitute the values of $a$ and $b$ into the formula:

Constant $= \frac{(\frac{3}{4})^2}{4 \times 9}$

Calculate the square of $\frac{3}{4}$:

$(\frac{3}{4})^2 = \frac{3^2}{4^2} = \frac{9}{16}$

Substitute this back into the expression for the constant:

Constant $= \frac{\frac{9}{16}}{36}$

Divide the fraction by 36:

Constant $= \frac{9}{16} \times \frac{1}{36}$

Cancel out common factors. The numerator 9 and the denominator 36 have a common factor of 9:

$\frac{\cancel{9}^{1}}{16} \times \frac{1}{\cancel{36}_{4}}$

Constant $= \frac{1}{16} \times \frac{1}{4}$

Multiply the fractions:

Constant $= \frac{1 \times 1}{16 \times 4} = \frac{1}{64}$

So, the constant that must be added and subtracted is $\frac{1}{64}$.

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Comparing this value with the given options:

(A) $\frac{1}{8}$

(B) $\frac{1}{64}$

(C) $\frac{1}{4}$

(D) $\frac{9}{64}$

The calculated constant $\frac{1}{64}$ matches option (B).

The correct answer is (B) $\frac{1}{64}$.

Solution:

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The given quadratic equation is $2x^2 - \sqrt{5}x + 1 = 0$.

For a standard quadratic equation in the form $ax^2 + bx + c = 0$, the nature of the roots is determined by the value of the discriminant, denoted by $\Delta$ or $D$, which is calculated as $\Delta = b^2 - 4ac$.

The relationship between the discriminant and the nature of roots is as follows:

• If $\Delta > 0$, the equation has two distinct real roots.
• If $\Delta = 0$, the equation has two equal real roots.
• If $\Delta < 0$, the equation has no real roots (the roots are complex conjugates).

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In the given equation $2x^2 - \sqrt{5}x + 1 = 0$, we identify the coefficients by comparing it to the standard form $ax^2 + bx + c = 0$:

$a = 2$

$b = -\sqrt{5}$

$c = 1$

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Now, we calculate the discriminant $\Delta = b^2 - 4ac$ using these values:

$\Delta = (-\sqrt{5})^2 - 4(2)(1)$

Calculate $(-\sqrt{5})^2$: $(-\sqrt{5})^2 = (-1)^2 \times (\sqrt{5})^2 = 1 \times 5 = 5$.

Calculate $4(2)(1)$: $4 \times 2 \times 1 = 8$.

Substitute these values back into the discriminant formula:

$\Delta = 5 - 8$

$\Delta = -3$

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Since the discriminant $\Delta = -3$, which is less than 0 ($\Delta < 0$), the quadratic equation $2x^2 - \sqrt{5}x + 1 = 0$ has no real roots.

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Comparing our finding with the given options:

(A) two distinct real roots (Requires $\Delta > 0$)

(B) two equal real roots (Requires $\Delta = 0$)

(C) no real roots (Requires $\Delta < 0$)

(D) more than 2 real roots (A quadratic equation has at most 2 roots)

Our result $\Delta = -3 < 0$ matches the condition for having no real roots.

The correct answer is (C) no real roots.

Solution:

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For a quadratic equation in the standard form $ax^2 + bx + c = 0$, the nature of the roots is determined by the discriminant, $\Delta = b^2 - 4ac$.

An equation has two distinct real roots if and only if its discriminant is strictly greater than zero ($\Delta > 0$).

We will calculate the discriminant for each given equation and check which one is positive.

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(A) $2x^2 – 3\sqrt{2}x + \frac{9}{4} = 0$

Here, $a = 2$, $b = -3\sqrt{2}$, $c = \frac{9}{4}$.

Discriminant $\Delta = b^2 - 4ac = (-3\sqrt{2})^2 - 4(2)(\frac{9}{4})$

Calculate $(-3\sqrt{2})^2 = (-3)^2 \times (\sqrt{2})^2 = 9 \times 2 = 18$.

Calculate $4(2)(\frac{9}{4}) = \cancel{4}^{1} \times 2 \times \frac{9}{\cancel{4}_{1}} = 2 \times 9 = 18$.

$\Delta = 18 - 18 = 0$.

Since $\Delta = 0$, this equation has two equal real roots, not two distinct real roots.

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(B) $x^2 + x – 5 = 0$

Here, $a = 1$, $b = 1$, $c = -5$.

Discriminant $\Delta = b^2 - 4ac = (1)^2 - 4(1)(-5)$

$\Delta = 1 - (-20)$

$\Delta = 1 + 20 = 21$.

Since $\Delta = 21 > 0$, this equation has two distinct real roots.

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(C) $x^2 + 3x + 2\sqrt{2} = 0$

Here, $a = 1$, $b = 3$, $c = 2\sqrt{2}$.

Discriminant $\Delta = b^2 - 4ac = (3)^2 - 4(1)(2\sqrt{2})$

$\Delta = 9 - 8\sqrt{2}$.

To determine if $9 - 8\sqrt{2}$ is positive or negative, we can compare $9$ with $8\sqrt{2}$.

Compare $9^2$ with $(8\sqrt{2})^2$:

$9^2 = 81$

$(8\sqrt{2})^2 = 8^2 \times (\sqrt{2})^2 = 64 \times 2 = 128$.

Since $81 < 128$, it means $9 < 8\sqrt{2}$.

Therefore, $9 - 8\sqrt{2} < 0$.

Since $\Delta < 0$, this equation has no real roots, not two distinct real roots.

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(D) $5x^2 – 3x + 1 = 0$

Here, $a = 5$, $b = -3$, $c = 1$.

Discriminant $\Delta = b^2 - 4ac = (-3)^2 - 4(5)(1)$

$\Delta = 9 - 20 = -11$.

Since $\Delta = -11 < 0$, this equation has no real roots, not two distinct real roots.

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Based on the discriminant values, only option (B) has a positive discriminant, indicating two distinct real roots.

The correct answer is (B) x² + x – 5 = 0.

Solution:

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For a quadratic equation in the standard form $ax^2 + bx + c = 0$, the nature of the roots is determined by the discriminant, $\Delta = b^2 - 4ac$.

An equation has no real roots if and only if its discriminant is strictly less than zero ($\Delta < 0$).

We will calculate the discriminant for each given equation and check which one is negative.

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(A) $x^2 – 4x + 3\sqrt{2} = 0$

Here, $a = 1$, $b = -4$, $c = 3\sqrt{2}$.

Discriminant $\Delta = b^2 - 4ac = (-4)^2 - 4(1)(3\sqrt{2})$

$\Delta = 16 - 12\sqrt{2}$.

To determine if $16 - 12\sqrt{2}$ is positive or negative, we compare $16$ with $12\sqrt{2}$.

Compare $16^2$ with $(12\sqrt{2})^2$:

$16^2 = 256$

$(12\sqrt{2})^2 = 12^2 \times (\sqrt{2})^2 = 144 \times 2 = 288$.

Since $256 < 288$, it means $16 < 12\sqrt{2}$.

Therefore, $16 - 12\sqrt{2} < 0$.

Since $\Delta < 0$, this equation has no real roots.

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(B) $x^2 + 4x – 3\sqrt{2} = 0$

Here, $a = 1$, $b = 4$, $c = -3\sqrt{2}$.

Discriminant $\Delta = b^2 - 4ac = (4)^2 - 4(1)(-3\sqrt{2})$

$\Delta = 16 - (-12\sqrt{2})$

$\Delta = 16 + 12\sqrt{2}$.

Since both terms are positive, $16 + 12\sqrt{2} > 0$.

Since $\Delta > 0$, this equation has two distinct real roots, not no real roots.

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(C) $x^2 – 4x – 3\sqrt{2} = 0$

Here, $a = 1$, $b = -4$, $c = -3\sqrt{2}$.

Discriminant $\Delta = b^2 - 4ac = (-4)^2 - 4(1)(-3\sqrt{2})$

$\Delta = 16 - (-12\sqrt{2})$

$\Delta = 16 + 12\sqrt{2}$.

Since both terms are positive, $16 + 12\sqrt{2} > 0$.

Since $\Delta > 0$, this equation has two distinct real roots, not no real roots.

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(D) $3x^2 + 4\sqrt{3} x + 4 = 0$

Here, $a = 3$, $b = 4\sqrt{3}$, $c = 4$.

Discriminant $\Delta = b^2 - 4ac = (4\sqrt{3})^2 - 4(3)(4)$

Calculate $(4\sqrt{3})^2 = 4^2 \times (\sqrt{3})^2 = 16 \times 3 = 48$.

Calculate $4(3)(4) = 4 \times 3 \times 4 = 48$.

$\Delta = 48 - 48 = 0$.

Since $\Delta = 0$, this equation has two equal real roots, not no real roots.

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Based on the discriminant values, only option (A) has a negative discriminant, indicating no real roots.

The correct answer is (A) x² – 4x + 3$\sqrt{2}$ = 0.

Solution:

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The given equation is $(x^2 + 1)^2 – x^2 = 0$.

First, we need to expand and simplify the equation to determine its form.

Expand the term $(x^2 + 1)^2$ using the identity $(a+b)^2 = a^2 + 2ab + b^2$:

$(x^2 + 1)^2 = (x^2)^2 + 2(x^2)(1) + 1^2 = x^4 + 2x^2 + 1$

Substitute this back into the original equation:

$x^4 + 2x^2 + 1 – x^2 = 0$

Combine the like terms ($2x^2$ and $-x^2$):

$x^4 + x^2 + 1 = 0$

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This is a biquadratic equation. Notice that it only contains terms with even powers of $x$. We can use a substitution to simplify it.

Let $y = x^2$. Since $x$ is a real number, $x^2$ must be non-negative, i.e., $y \geq 0$.

Substituting $y = x^2$ (and $x^4 = (x^2)^2 = y^2$) into the simplified equation $x^4 + x^2 + 1 = 0$, we get:

$y^2 + y + 1 = 0$

This is a quadratic equation in the variable $y$. To find the nature of the roots for $y$, we calculate its discriminant $\Delta_y = B^2 - 4AC$, where $A=1$, $B=1$, and $C=1$ are the coefficients of the quadratic in $y$.

$\Delta_y = (1)^2 - 4(1)(1)$

$\Delta_y = 1 - 4$

$\Delta_y = -3$

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Since the discriminant $\Delta_y = -3$ is less than zero ($\Delta_y < 0$), the quadratic equation $y^2 + y + 1 = 0$ has no real roots for $y$. The roots for $y$ are complex numbers.

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Recall that we made the substitution $y = x^2$. For a real number $x$, $x^2$ must be a non-negative real number ($x^2 \geq 0$).

Since the quadratic equation in $y$ ($y^2 + y + 1 = 0$) has no real solutions for $y$ (let alone non-negative real solutions), there are no real values of $x$ for which $x^2$ can satisfy this equation.

Therefore, the original equation $(x^2 + 1)^2 – x^2 = 0$ has no real roots.

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Comparing our conclusion with the given options:

(A) four real roots

(B) two real roots

(C) no real roots

(D) one real root

Our finding matches option (C).

The correct answer is (C) no real roots.

Solution:

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The given equation is $(x – 1)^2 + 2(x + 1) = 0$.

To determine if the equation has real roots, we need to simplify it into the standard quadratic form $ax^2 + bx + c = 0$ and then calculate its discriminant.

First, expand the terms in the equation:

$(x – 1)^2 = x^2 - 2(x)(1) + 1^2 = x^2 - 2x + 1$

$2(x + 1) = 2x + 2$

Substitute these expanded forms back into the original equation:

$(x^2 - 2x + 1) + (2x + 2) = 0$

Combine like terms:

$x^2 + (-2x + 2x) + (1 + 2) = 0$

$x^2 + 0x + 3 = 0$

The simplified equation is $x^2 + 3 = 0$.

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Justification:

For a quadratic equation in the form $ax^2 + bx + c = 0$, the nature of the roots is determined by the discriminant $\Delta = b^2 - 4ac$.

• If $\Delta > 0$, there are two distinct real roots.
• If $\Delta = 0$, there are two equal real roots.
• If $\Delta < 0$, there are no real roots.

From the simplified equation $x^2 + 0x + 3 = 0$, we have the coefficients:

$a = 1$

$b = 0$

$c = 3$

Calculate the discriminant:

$\Delta = b^2 - 4ac = (0)^2 - 4(1)(3)$

$\Delta = 0 - 12$

$\Delta = -12$

Since the discriminant $\Delta = -12$ is less than zero ($\Delta < 0$), the quadratic equation $x^2 + 3 = 0$ has no real roots.

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Conclusion:

No, the equation $(x – 1)^2 + 2(x + 1) = 0$ does not have a real root. This is because its discriminant is $-12$, which is less than zero.

Statement:

"If in a quadratic equation the coefficient of $x$ is zero, then the quadratic equation has no real roots."

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Solution:

Let the standard form of a quadratic equation be $ax^2 + bx + c = 0$, where $a \neq 0$.

The statement says that the coefficient of $x$ is zero, which means $b = 0$.

So, the quadratic equation becomes $ax^2 + 0x + c = 0$, which simplifies to $ax^2 + c = 0$.

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Justification:

The nature of the roots of a quadratic equation is determined by its discriminant, $\Delta = b^2 - 4ac$.

For real roots to exist, the discriminant must be greater than or equal to zero ($\Delta \geq 0$). For no real roots to exist, the discriminant must be less than zero ($\Delta < 0$).

In the case where the coefficient of $x$ is zero ($b=0$), the discriminant is calculated as:

$\Delta = (0)^2 - 4ac$

$\Delta = -4ac$

For the equation to have no real roots, we require $\Delta < 0$.

$-4ac < 0$

Dividing both sides by $-4$ (and reversing the inequality sign):

$ac > 0$

This means that the equation $ax^2 + c = 0$ has no real roots only if the coefficients $a$ and $c$ have the same sign (both positive or both negative).

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However, if $a$ and $c$ have opposite signs, then $ac < 0$, which means $-4ac > 0$. In this case, $\Delta > 0$, and the equation has two distinct real roots.

For example, consider the equation $x^2 - 4 = 0$. Here, $a=1$, $b=0$, $c=-4$. The coefficient of $x$ is zero. The discriminant is $\Delta = (0)^2 - 4(1)(-4) = 16$. Since $\Delta = 16 > 0$, this equation has two distinct real roots ($x = 2$ and $x = -2$).

Another example is $x^2 = 0$. Here, $a=1$, $b=0$, $c=0$. The coefficient of $x$ is zero. The discriminant is $\Delta = (0)^2 - 4(1)(0) = 0$. Since $\Delta = 0$, this equation has two equal real roots ($x = 0$ repeated).

These examples demonstrate that a quadratic equation with the coefficient of $x$ being zero can indeed have real roots.

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Conclusion:

The statement "If in a quadratic equation the coefficient of $x$ is zero, then the quadratic equation has no real roots" is False.

Justification: A quadratic equation $ax^2 + c = 0$ (where $b=0$) has real roots if $ac \leq 0$. This occurs when $a$ and $c$ have opposite signs or when $c=0$. Only when $a$ and $c$ have the same sign ($ac > 0$) does the equation have no real roots.

Solution:

A quadratic equation in the standard form $ax^2 + bx + c = 0$ ($a \neq 0$) has two distinct real roots if and only if its discriminant, $\Delta = b^2 - 4ac$, is strictly greater than zero ($\Delta > 0$). We will calculate the discriminant for each equation and check its sign.

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(i) $x^2 – 3x + 4 = 0$

Comparing with $ax^2 + bx + c = 0$, we have $a = 1$, $b = -3$, $c = 4$.

Discriminant $\Delta = b^2 - 4ac = (-3)^2 - 4(1)(4)$

$\Delta = 9 - 16$

$\Delta = -7$

Since $\Delta = -7 < 0$, the equation has no real roots.

Conclusion: No. Justification: The discriminant is negative ($\Delta = -7$).

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(ii) $2x^2 + x – 1 = 0$

Comparing with $ax^2 + bx + c = 0$, we have $a = 2$, $b = 1$, $c = -1$.

Discriminant $\Delta = b^2 - 4ac = (1)^2 - 4(2)(-1)$

$\Delta = 1 - (-8)$

$\Delta = 1 + 8 = 9$

Since $\Delta = 9 > 0$, the equation has two distinct real roots.

Conclusion: Yes. Justification: The discriminant is positive ($\Delta = 9$).

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(iii) $2x^2 – 6x + \frac{9}{2} = 0$

Comparing with $ax^2 + bx + c = 0$, we have $a = 2$, $b = -6$, $c = \frac{9}{2}$.

Discriminant $\Delta = b^2 - 4ac = (-6)^2 - 4(2)(\frac{9}{2})$

$\Delta = 36 - \cancel{4}^{2} \times \cancel{2}^{1} \times \frac{9}{\cancel{2}_{1}}$

$\Delta = 36 - 36 = 0$

Since $\Delta = 0$, the equation has two equal real roots.

Conclusion: No. Justification: The discriminant is zero ($\Delta = 0$).

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(iv) $3x^2 – 4x + 1 = 0$

Comparing with $ax^2 + bx + c = 0$, we have $a = 3$, $b = -4$, $c = 1$.

Discriminant $\Delta = b^2 - 4ac = (-4)^2 - 4(3)(1)$

$\Delta = 16 - 12$

$\Delta = 4$

Since $\Delta = 4 > 0$, the equation has two distinct real roots.

Conclusion: Yes. Justification: The discriminant is positive ($\Delta = 4$).

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(v) $(x + 4)^2 – 8x = 0$

First, simplify the equation:

$(x^2 + 8x + 16) - 8x = 0$

$x^2 + 16 = 0$

Comparing with $ax^2 + bx + c = 0$, we have $a = 1$, $b = 0$, $c = 16$.

Discriminant $\Delta = b^2 - 4ac = (0)^2 - 4(1)(16)$

$\Delta = 0 - 64$

$\Delta = -64$

Since $\Delta = -64 < 0$, the equation has no real roots.

Conclusion: No. Justification: The discriminant is negative ($\Delta = -64$).

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(vi) $(x – \sqrt{2})^2 – 2(x + 1) = 0$

First, simplify the equation:

$(x^2 - 2\sqrt{2}x + (\sqrt{2})^2) - (2x + 2) = 0$

$x^2 - 2\sqrt{2}x + 2 - 2x - 2 = 0$

$x^2 + (-2\sqrt{2} - 2)x + (2 - 2) = 0$

$x^2 - (2\sqrt{2} + 2)x + 0 = 0$

Comparing with $ax^2 + bx + c = 0$, we have $a = 1$, $b = -(2\sqrt{2} + 2)$, $c = 0$.

Discriminant $\Delta = b^2 - 4ac = (-(2\sqrt{2} + 2))^2 - 4(1)(0)$

$\Delta = (2\sqrt{2} + 2)^2 - 0$

$\Delta = (2\sqrt{2})^2 + 2(2\sqrt{2})(2) + 2^2$

$\Delta = (4 \times 2) + 8\sqrt{2} + 4$

$\Delta = 8 + 8\sqrt{2} + 4 = 12 + 8\sqrt{2}$

Since $\sqrt{2}$ is positive, $8\sqrt{2}$ is positive, and $12$ is positive, $\Delta = 12 + 8\sqrt{2} > 0$.

Since $\Delta > 0$, the equation has two distinct real roots.

Conclusion: Yes. Justification: The discriminant is positive ($\Delta = 12 + 8\sqrt{2}$).

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(vii) $\sqrt{2}$x² - $\frac{3}{\sqrt{2}}$x + $\frac{1}{\sqrt{2}}$ = 0

Comparing with $ax^2 + bx + c = 0$, we have $a = \sqrt{2}$, $b = -\frac{3}{\sqrt{2}}$, $c = \frac{1}{\sqrt{2}}$.

Discriminant $\Delta = b^2 - 4ac = (-\frac{3}{\sqrt{2}})^2 - 4(\sqrt{2})(\frac{1}{\sqrt{2}})$

$\Delta = \frac{(-3)^2}{(\sqrt{2})^2} - 4(\cancel{\sqrt{2}})(\frac{1}{\cancel{\sqrt{2}}})$

$\Delta = \frac{9}{2} - 4(1)$

$\Delta = \frac{9}{2} - 4 = \frac{9}{2} - \frac{8}{2} = \frac{1}{2}$

Since $\Delta = \frac{1}{2} > 0$, the equation has two distinct real roots.

Conclusion: Yes. Justification: The discriminant is positive ($\Delta = \frac{1}{2}$).

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(viii) $x (1 – x) – 2 = 0$

First, simplify the equation:

$x - x^2 - 2 = 0$

Rearrange in standard form: $-x^2 + x - 2 = 0$

Comparing with $ax^2 + bx + c = 0$, we have $a = -1$, $b = 1$, $c = -2$.

Discriminant $\Delta = b^2 - 4ac = (1)^2 - 4(-1)(-2)$

$\Delta = 1 - (8)$

$\Delta = 1 - 8 = -7$

Since $\Delta = -7 < 0$, the equation has no real roots.

Conclusion: No. Justification: The discriminant is negative ($\Delta = -7$).

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(ix) $(x – 1) (x + 2) + 2 = 0$

First, simplify the equation:

$x(x + 2) - 1(x + 2) + 2 = 0$

$x^2 + 2x - x - 2 + 2 = 0$

$x^2 + x = 0$

Comparing with $ax^2 + bx + c = 0$, we have $a = 1$, $b = 1$, $c = 0$.

Discriminant $\Delta = b^2 - 4ac = (1)^2 - 4(1)(0)$

$\Delta = 1 - 0$

$\Delta = 1$

Since $\Delta = 1 > 0$, the equation has two distinct real roots.

Conclusion: Yes. Justification: The discriminant is positive ($\Delta = 1$).

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(x) $(x + 1) (x – 2) + x = 0$

First, simplify the equation:

$x(x – 2) + 1(x – 2) + x = 0$

$x^2 - 2x + x - 2 + x = 0$

$x^2 + (-2x + x + x) - 2 = 0$

$x^2 + 0x - 2 = 0$

Comparing with $ax^2 + bx + c = 0$, we have $a = 1$, $b = 0$, $c = -2$.

Discriminant $\Delta = b^2 - 4ac = (0)^2 - 4(1)(-2)$

$\Delta = 0 - (-8)$

$\Delta = 8$

Since $\Delta = 8 > 0$, the equation has two distinct real roots.

Conclusion: Yes. Justification: The discriminant is positive ($\Delta = 8$).

Solution:

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A quadratic equation in the standard form is $ax^2 + bx + c = 0$, where $a \neq 0$. The nature and number of roots depend on the discriminant $\Delta = b^2 - 4ac$.

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(i) Every quadratic equation has exactly one root.

False. A quadratic equation can have two distinct real roots (if $\Delta > 0$), two equal real roots (if $\Delta = 0$), or two distinct complex roots (if $\Delta < 0$). It never has exactly one root (unless we consider the case of equal roots as counting for two roots).

Justification: Consider $x^2 - 4 = 0$. The roots are $x = 2$ and $x = -2$ (two distinct real roots). Consider $x^2 = 0$. The root is $x = 0$ with multiplicity 2 (two equal real roots). Consider $x^2 + 1 = 0$. The roots are $x = i$ and $x = -i$ (two distinct complex roots). None of these examples have exactly one root.

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(ii) Every quadratic equation has at least one real root.

False. A quadratic equation can have complex roots (when $\Delta < 0$). These roots are not real.

Justification: Consider $x^2 + 1 = 0$. Here $a=1, b=0, c=1$. The discriminant is $\Delta = 0^2 - 4(1)(1) = -4$. Since $\Delta < 0$, the roots are not real. The roots are $x = \pm i$.

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(iii) Every quadratic equation has at least two roots.

True. According to the Fundamental Theorem of Algebra, a polynomial equation of degree $n$ has exactly $n$ roots in the complex number system (counting multiplicity). A quadratic equation is a polynomial of degree 2, so it has exactly 2 roots in the complex number system. These roots can be real or complex, distinct or equal.

Justification: The quadratic formula $\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ shows that there are always two values for $x$ (corresponding to the $\pm$ sign), unless the discriminant is zero, in which case the two values are equal. In the complex number system, $\sqrt{b^2 - 4ac}$ is always defined, even if $b^2 - 4ac$ is negative.

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(iv) Every quadratic equation has at most two roots.

True. This is a direct consequence of the fact that a quadratic equation is a polynomial of degree 2. A polynomial of degree $n$ has exactly $n$ roots when counted with multiplicity in the complex numbers, and at most $n$ distinct roots.

Justification: If a quadratic equation had more than two roots, say $r_1, r_2, r_3$ are distinct roots, then the polynomial $ax^2 + bx + c$ would have factors $(x-r_1)$, $(x-r_2)$, and $(x-r_3)$. The product $(x-r_1)(x-r_2)(x-r_3)$ is a cubic polynomial, not a quadratic polynomial (unless $a=0$, which is not allowed for a quadratic equation). Therefore, a quadratic equation cannot have more than two distinct roots. Counting multiplicity, it has exactly two roots.

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(v) If the coefficient of x² and the constant term of a quadratic equation have opposite signs, then the quadratic equation has real roots.

True. Let the equation be $ax^2 + bx + c = 0$. The coefficient of $x^2$ is $a$, and the constant term is $c$. The statement says that $a$ and $c$ have opposite signs, which means their product $ac < 0$.

The discriminant is $\Delta = b^2 - 4ac$.

Since $ac < 0$, the term $-4ac$ is positive (since $-4 \times \text{negative number} = \text{positive number}$).

So, $\Delta = b^2 + (\text{a positive number})$.

Since $b^2 \geq 0$ and $-4ac > 0$, their sum $b^2 - 4ac$ must be strictly positive.

$\Delta > 0$

Since the discriminant is positive, the equation has two distinct real roots. Therefore, it has real roots.

Justification: If $a$ and $c$ have opposite signs, then $ac < 0$. The discriminant is $\Delta = b^2 - 4ac$. Since $b^2 \geq 0$ and $-4ac > 0$, we have $\Delta = b^2 + (-4ac) > 0$. A positive discriminant guarantees real roots.

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(vi) If the coefficient of x² and the constant term have the same sign and if the coefficient of x term is zero, then the quadratic equation has no real roots.

True. Let the equation be $ax^2 + bx + c = 0$. The coefficient of $x^2$ is $a$, the constant term is $c$, and the coefficient of $x$ is $b$. The statement says $a$ and $c$ have the same sign, and $b = 0$.

Since $a$ and $c$ have the same sign, their product $ac > 0$.

The equation becomes $ax^2 + 0x + c = 0$, or $ax^2 + c = 0$.

The discriminant is $\Delta = b^2 - 4ac$. Substitute $b=0$:

$\Delta = (0)^2 - 4ac$

$\Delta = -4ac$

Since $ac > 0$, the term $-4ac$ is negative (since $-4 \times \text{positive number} = \text{negative number}$).

So, $\Delta < 0$.

Since the discriminant is negative, the equation has no real roots.

Justification: If $a$ and $c$ have the same sign, $ac > 0$. If $b=0$, the discriminant is $\Delta = 0^2 - 4ac = -4ac$. Since $ac > 0$, $-4ac < 0$. A negative discriminant means no real roots.

Statement:

"A quadratic equation with integral coefficients has integral roots."

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Solution:

Let the quadratic equation be $ax^2 + bx + c = 0$, where $a, b, c$ are integers and $a \neq 0$.

The roots of this equation are given by the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

For the roots to be integral, two conditions must be met:

1. The discriminant, $\Delta = b^2 - 4ac$, must be a perfect square of an integer (or zero). If $\Delta$ is not a perfect square, the square root $\sqrt{\Delta}$ will be irrational, and the roots will be irrational (unless $b=0$ and $a$ is a perfect square times a common factor with $c$, which is not generally true).
2. The numerator, $-b \pm \sqrt{\Delta}$, must be divisible by the denominator, $2a$.

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Justification (by counterexample):

Consider the quadratic equation $x^2 - 3x + 2 = 0$.

Here, the coefficients are integers: $a=1$, $b=-3$, $c=2$.

The roots can be found by factoring: $(x-1)(x-2) = 0$. The roots are $x = 1$ and $x = 2$. These are integers.

In this case, the statement holds true.

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Now, consider the quadratic equation $2x^2 - 3x + 1 = 0$.

Here, the coefficients are integers: $a=2$, $b=-3$, $c=1$.

We can find the roots using the quadratic formula:

$\Delta = b^2 - 4ac = (-3)^2 - 4(2)(1) = 9 - 8 = 1$

$x = \frac{-(-3) \pm \sqrt{1}}{2(2)}$

$x = \frac{3 \pm 1}{4}$

The roots are $x_1 = \frac{3 + 1}{4} = \frac{4}{4} = 1$ and $x_2 = \frac{3 - 1}{4} = \frac{2}{4} = \frac{1}{2}$.

One root is $1$ (an integer), but the other root is $\frac{1}{2}$ (not an integer).

This provides a counterexample to the statement.

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Another counterexample:

Consider the equation $x^2 - x - 1 = 0$.

Here, coefficients are integers: $a=1, b=-1, c=-1$.

$\Delta = b^2 - 4ac = (-1)^2 - 4(1)(-1) = 1 + 4 = 5$.

The roots are $x = \frac{-(-1) \pm \sqrt{5}}{2(1)} = \frac{1 \pm \sqrt{5}}{2}$.

Since $\sqrt{5}$ is irrational, the roots $\frac{1 + \sqrt{5}}{2}$ and $\frac{1 - \sqrt{5}}{2}$ are irrational numbers, which are not integers.

This is another counterexample.

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Conclusion:

The statement "A quadratic equation with integral coefficients has integral roots" is False.

Justification: While some quadratic equations with integral coefficients have integral roots (e.g., $x^2 - 3x + 2 = 0$ with roots 1 and 2), it is not always true. For example, the equation $2x^2 - 3x + 1 = 0$ has integral coefficients but one root is $\frac{1}{2}$, which is not an integer. The equation $x^2 - x - 1 = 0$ has integral coefficients but irrational roots ($\frac{1 \pm \sqrt{5}}{2}$), which are not integers.

Statement:

"Does there exist a quadratic equation whose coefficients are rational but both of its roots are irrational?"

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Solution:

Let the quadratic equation be $ax^2 + bx + c = 0$, where $a, b, c$ are rational numbers and $a \neq 0$.

The roots of the equation are given by the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

For the roots to be irrational, the discriminant, $\Delta = b^2 - 4ac$, must be positive but not a perfect square of a rational number. Also, the coefficients $a$, $b$, and $c$ are rational.

Since $a$, $b$, and $c$ are rational, their squares and products are also rational. Thus, $b^2$ is rational, $4ac$ is rational, and their difference $b^2 - 4ac$ is rational.

Let $\Delta = k$, where $k$ is a rational number. The roots are $x = \frac{-b \pm \sqrt{k}}{2a}$.

For the roots to be irrational, $\sqrt{k}$ must be irrational. This happens if $k$ is a positive rational number that is not the square of a rational number.

So, we need to find a quadratic equation with rational coefficients such that its discriminant is a positive rational number that is not a perfect square of a rational number.

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Justification (by example):

Consider the quadratic equation $x^2 - 2x - 1 = 0$.

Here, the coefficients are $a=1$, $b=-2$, $c=-1$. These are all integers, and thus they are rational numbers.

Calculate the discriminant $\Delta = b^2 - 4ac$:

$\Delta = (-2)^2 - 4(1)(-1)$

$\Delta = 4 - (-4)$

$\Delta = 4 + 4 = 8$

The discriminant $\Delta = 8$ is a positive rational number.

Now, consider $\sqrt{\Delta} = \sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2}$.

$2\sqrt{2}$ is an irrational number because $\sqrt{2}$ is irrational.

The roots of the equation are $x = \frac{-b \pm \sqrt{\Delta}}{2a} = \frac{-(-2) \pm \sqrt{8}}{2(1)} = \frac{2 \pm 2\sqrt{2}}{2} = 1 \pm \sqrt{2}$.

The roots are $x_1 = 1 + \sqrt{2}$ and $x_2 = 1 - \sqrt{2}$.

Both $1 + \sqrt{2}$ and $1 - \sqrt{2}$ are irrational numbers.

Thus, we have found a quadratic equation ($x^2 - 2x - 1 = 0$) whose coefficients are rational (1, -2, -1) and whose roots are both irrational ($1 + \sqrt{2}$ and $1 - \sqrt{2}$).

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Conclusion:

Yes, there exists a quadratic equation whose coefficients are rational but both of its roots are irrational.

Justification: As shown by the example $x^2 - 2x - 1 = 0$, which has rational coefficients ($a=1, b=-2, c=-1$) and irrational roots ($1 \pm \sqrt{2}$), such equations exist. This happens when the discriminant ($b^2 - 4ac$) is a positive rational number that is not a perfect square of a rational number, resulting in $\sqrt{b^2 - 4ac}$ being irrational.

Statement:

"Does there exist a quadratic equation whose coefficients are all distinct irrationals but both the roots are rationals?"

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Solution:

Let the quadratic equation be $ax^2 + bx + c = 0$, where $a, b, c$ are distinct irrational numbers and $a \neq 0$.

Let the roots of the equation be $r_1$ and $r_2$, where $r_1$ and $r_2$ are rational numbers.

From Vieta's formulas, for a quadratic equation $ax^2 + bx + c = 0$, the sum of the roots is $r_1 + r_2 = -\frac{b}{a}$ and the product of the roots is $r_1 r_2 = \frac{c}{a}$.

Since $r_1$ and $r_2$ are rational, their sum $r_1 + r_2$ is rational, and their product $r_1 r_2$ is rational.

So, we have:

From the first equation, since $-\frac{b}{a}$ is rational, $\frac{b}{a}$ is also rational. Let $\frac{b}{a} = q_1$, where $q_1$ is a rational number.

So, $b = q_1 a$.

From the second equation, $\frac{c}{a} = q_2$, where $q_2$ is a rational number.

So, $c = q_2 a$.

We are given that $a, b, c$ are distinct irrational numbers.

We have $b = q_1 a$ and $c = q_2 a$, where $q_1$ and $q_2$ are rational.

If $a$ is an irrational number, then for $b = q_1 a$ to be an irrational number, $q_1$ must be non-zero. Similarly, for $c = q_2 a$ to be an irrational number, $q_2$ must be non-zero.

Also, for $b$ and $c$ to be distinct from $a$, we must have $q_1 \neq 1$ and $q_2 \neq 1$.

For $b$ and $c$ to be distinct from each other, we must have $q_1 a \neq q_2 a$, which means $q_1 \neq q_2$ (since $a \neq 0$).

So, we need to find distinct non-zero rational numbers $q_1, q_2 \neq 1$ and an irrational number $a$ such that $a$, $q_1 a$, and $q_2 a$ are distinct irrational numbers.

Let's try to construct such an equation. Suppose the roots are $r_1 = 1$ and $r_2 = 2$. These are rational and distinct.

Sum of roots: $1 + 2 = 3$. So, $-\frac{b}{a} = 3$, which implies $b = -3a$.

Product of roots: $1 \times 2 = 2$. So, $\frac{c}{a} = 2$, which implies $c = 2a$.

We need $a$, $b$, and $c$ to be distinct irrational numbers.

Let $a = \sqrt{2}$. This is an irrational number.

Then $b = -3a = -3\sqrt{2}$. This is an irrational number.

And $c = 2a = 2\sqrt{2}$. This is an irrational number.

Check if $a, b, c$ are distinct:

$a = \sqrt{2}$

$b = -3\sqrt{2}$

$c = 2\sqrt{2}$

These three numbers ($\sqrt{2}$, $-3\sqrt{2}$, $2\sqrt{2}$) are indeed distinct irrational numbers.

The quadratic equation is $\sqrt{2}x^2 - 3\sqrt{2}x + 2\sqrt{2} = 0$.

The coefficients are $a = \sqrt{2}$, $b = -3\sqrt{2}$, $c = 2\sqrt{2}$, which are distinct irrational numbers.

The roots are 1 and 2, which are rational numbers.

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Conclusion:

Yes, such a quadratic equation exists.

Justification: We can construct such an equation using Vieta's formulas. If the roots are rational numbers $r_1$ and $r_2$, then the sum $r_1 + r_2$ and the product $r_1 r_2$ are rational. We have $-\frac{b}{a} = r_1 + r_2$ and $\frac{c}{a} = r_1 r_2$. We can choose an irrational number for $a$ (e.g., $\sqrt{2}$) and distinct rational numbers for the roots (e.g., 1 and 2). Then we can solve for $b$ and $c$ to be irrational numbers distinct from $a$ and from each other. In our example with roots 1 and 2, if we choose $a = \sqrt{2}$, then $b = -(r_1+r_2)a = -3\sqrt{2}$ and $c = (r_1 r_2)a = 2\sqrt{2}$. The coefficients $\sqrt{2}$, $-3\sqrt{2}$, and $2\sqrt{2}$ are distinct irrational numbers, and the roots are rational (1 and 2).

Statement:

"Is 0.2 a root of the equation $x^2 – 0.4 = 0$?"

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Solution:

A number is a root of an equation if substituting the number for the variable makes the equation true.

We need to check if substituting $x = 0.2$ into the equation $x^2 – 0.4 = 0$ satisfies the equation.

Substitute $x = 0.2$ into the left side of the equation:

Left Side $= (0.2)^2 - 0.4$

Calculate $(0.2)^2$:

$(0.2)^2 = 0.2 \times 0.2 = 0.04$

Now, substitute this value back into the expression:

Left Side $= 0.04 - 0.4$

Subtract 0.4 from 0.04:

$0.04 - 0.40 = -0.36$

The left side evaluates to $-0.36$.

The right side of the equation is 0.

Since $-0.36 \neq 0$, the equation is not satisfied when $x = 0.2$.

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Justification:

To check if $0.2$ is a root of $x^2 - 0.4 = 0$, we substitute $x = 0.2$ into the equation:

$= 0.04 - 0.4$

$= -0.36$

Since the result $-0.36$ is not equal to $0$, $x = 0.2$ does not satisfy the equation.

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Alternatively, we can find the actual roots of the equation $x^2 - 0.4 = 0$ and see if $0.2$ is one of them.

$x^2 = 0.4$

$x = \pm \sqrt{0.4}$

$x = \pm \sqrt{\frac{4}{10}} = \pm \sqrt{\frac{2}{5}} = \pm \frac{\sqrt{2}}{\sqrt{5}} = \pm \frac{\sqrt{10}}{5}$

The roots are $\frac{\sqrt{10}}{5}$ and $-\frac{\sqrt{10}}{5}$.

We know that $\sqrt{9} = 3$ and $\sqrt{16} = 4$, so $\sqrt{10}$ is slightly larger than 3, approximately 3.16.

$\frac{\sqrt{10}}{5} \approx \frac{3.16}{5} = 0.632$

The roots are approximately $0.632$ and $-0.632$.

Since $0.2$ is not equal to $0.632$ or $-0.632$, $0.2$ is not a root of the equation.

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Conclusion:

No, 0.2 is not a root of the equation $x^2 – 0.4 = 0$.

Justification: Substituting $x=0.2$ into the equation yields $(0.2)^2 - 0.4 = 0.04 - 0.4 = -0.36$, which is not equal to 0. Therefore, 0.2 does not satisfy the equation and is not a root.

Statement:

"If $b = 0$ and $c < 0$, the roots of $x^2 + bx + c = 0$ are numerically equal and opposite in sign."

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Solution:

The given quadratic equation is $x^2 + bx + c = 0$.

We are given the conditions $b = 0$ and $c < 0$.

Substituting $b = 0$ into the equation, we get:

$x^2 + (0)x + c = 0$

$x^2 + c = 0$

Since $c < 0$, we can write $c = -k$ for some positive number $k$ (i.e., $k > 0$).

Substituting $c = -k$ into the equation:

$x^2 - k = 0$

Solve for $x^2$:

$x^2 = k$

Since $k > 0$, we can take the square root of both sides:

$x = \pm \sqrt{k}$

The two roots are $x_1 = \sqrt{k}$ and $x_2 = -\sqrt{k}$.

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Justification:

We have the roots as $\sqrt{k}$ and $-\sqrt{k}$, where $k = -c$ and $k > 0$.

• Real roots: Since $k > 0$, $\sqrt{k}$ is a real number. Thus, both roots are real.
• Numerically equal: The numerical value (or absolute value) of $\sqrt{k}$ is $|\sqrt{k}| = \sqrt{k}$. The numerical value of $-\sqrt{k}$ is $|-\sqrt{k}| = \sqrt{k}$. Since $\sqrt{k} = \sqrt{k}$, the roots are numerically equal.
• Opposite in sign: One root is $\sqrt{k}$ (which is positive since $k>0$) and the other root is $-\sqrt{k}$ (which is negative). Thus, the roots are opposite in sign.

For example, if $c = -4$, then $k = -c = 4$. The equation is $x^2 - 4 = 0$. The roots are $x = \pm \sqrt{4} = \pm 2$. The roots are 2 and -2, which are numerically equal (both have numerical value 2) and opposite in sign.

Since the roots are $x = \pm \sqrt{-c}$ and $c < 0$, $-c > 0$. Let $\sqrt{-c} = \alpha$, where $\alpha$ is a real number. The roots are $\alpha$ and $-\alpha$. These are clearly numerically equal ($|\alpha| = |-\alpha| = \alpha$) and opposite in sign.

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Conclusion:

Yes, the statement is True.

Justification: When $b=0$ and $c<0$, the quadratic equation simplifies to $x^2 = -c$. Since $c<0$, $-c > 0$. The roots are $x = \pm \sqrt{-c}$. Let $\sqrt{-c} = \alpha$. The roots are $\alpha$ and $-\alpha$. These roots are real, numerically equal ($|\alpha| = |-\alpha|$), and opposite in sign.

Solution:

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The given quadratic equation is $2x^2 – \sqrt{5}x – 2 = 0$.

We need to find the roots using the quadratic formula.

The standard form of a quadratic equation is $ax^2 + bx + c = 0$. Comparing the given equation with the standard form, we identify the coefficients:

$a = 2$

$b = -\sqrt{5}$

$c = -2$

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The quadratic formula for finding the roots of $ax^2 + bx + c = 0$ is:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

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First, calculate the discriminant $\Delta = b^2 - 4ac$:

$\Delta = (-\sqrt{5})^2 - 4(2)(-2)$

$\Delta = 5 - (-16)$

$\Delta = 5 + 16 = 21$

Since the discriminant $\Delta = 21 > 0$, the equation has two distinct real roots.

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Now, substitute the values of $a$, $b$, and $\Delta$ into the quadratic formula:

$x = \frac{-(-\sqrt{5}) \pm \sqrt{21}}{2(2)}$

$x = \frac{\sqrt{5} \pm \sqrt{21}}{4}$

The two roots are:

$x_1 = \frac{\sqrt{5} + \sqrt{21}}{4}$

$x_2 = \frac{\sqrt{5} - \sqrt{21}}{4}$

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The roots of the quadratic equation $2x^2 – \sqrt{5}x – 2 = 0$ are $\frac{\sqrt{5} + \sqrt{21}}{4}$ and $\frac{\sqrt{5} - \sqrt{21}}{4}$.

Solution:

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The given quadratic equation is:

$6x^2 - \sqrt{2}x - 2 = 0$

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To find the roots by the factorisation method, we need to split the middle term ($-\sqrt{2}x$) into two terms such that their sum is $-\sqrt{2}x$ and their product is equal to the product of the coefficient of $x^2$ (which is 6) and the constant term (which is -2).

Product of the coefficient of $x^2$ and the constant term $= (6) \times (-2) = -12$.

Coefficient of the middle term $= -\sqrt{2}$.

We need to find two numbers whose product is $-12$ and sum is $-\sqrt{2}$. Let these numbers be $p\sqrt{2}$ and $q\sqrt{2}$.

Their sum is $p\sqrt{2} + q\sqrt{2} = (p+q)\sqrt{2}$. Comparing this with $-\sqrt{2}$, we get $p+q = -1$.

Their product is $(p\sqrt{2})(q\sqrt{2}) = pq (\sqrt{2})^2 = 2pq$. Comparing this with $-12$, we get $2pq = -12$, which simplifies to $pq = -6$.

We need two numbers $p$ and $q$ such that $p+q = -1$ and $pq = -6$. These numbers are $2$ and $-3$.

So, the two terms for splitting the middle term are $2\sqrt{2}x$ and $-3\sqrt{2}x$. We can rewrite $-\sqrt{2}x$ as $2\sqrt{2}x - 3\sqrt{2}x$.

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Substitute the split middle term back into the equation:

$6x^2 + 2\sqrt{2}x - 3\sqrt{2}x - 2 = 0$

Group the terms:

$(6x^2 + 2\sqrt{2}x) - (3\sqrt{2}x + 2) = 0$

Factor out common terms from each group:

From the first group, $2x$ is common:

$6x^2 + 2\sqrt{2}x = 2x(3x + \sqrt{2})$

From the second group, note that $2$ can be written as $\sqrt{2} \times \sqrt{2}$. So $\sqrt{2}$ is common:

$3\sqrt{2}x + 2 = 3\sqrt{2}x + \sqrt{2}\sqrt{2} = \sqrt{2}(3x + \sqrt{2})$

Substitute the factored terms back into the equation:

$2x(3x + \sqrt{2}) - \sqrt{2}(3x + \sqrt{2}) = 0$

Factor out the common binomial term $(3x + \sqrt{2})$:

$(3x + \sqrt{2})(2x - \sqrt{2}) = 0$

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For the product of two factors to be zero, at least one of the factors must be zero.

Case 1:

$3x + \sqrt{2} = 0$

$3x = -\sqrt{2}$

$x = -\frac{\sqrt{2}}{3}$

Case 2:

$2x - \sqrt{2} = 0$

$2x = \sqrt{2}$

$x = \frac{\sqrt{2}}{2}$

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Therefore, the roots of the quadratic equation $6x^2 – \sqrt{2}x – 2 = 0$ are $x = -\frac{\sqrt{2}}{3}$ and $x = \frac{\sqrt{2}}{2}$.

The general form of a quadratic equation is $ax^2 + bx + c = 0$. The roots of a quadratic equation can be found using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

The discriminant is $\Delta = b^2 - 4ac$.

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(i) $2x^2 – 3x – 5 = 0$

Comparing with $ax^2 + bx + c = 0$, we have $a=2$, $b=-3$, $c=-5$.

Discriminant $\Delta = b^2 - 4ac = (-3)^2 - 4(2)(-5) = 9 + 40 = 49$.

Since $\Delta = 49 > 0$, the equation has two distinct real roots.

Using the quadratic formula:

$x = \frac{-(-3) \pm \sqrt{49}}{2(2)}$

$x = \frac{3 \pm 7}{4}$

The roots are:

$x_1 = \frac{3 + 7}{4} = \frac{10}{4} = \frac{5}{2}$

$x_2 = \frac{3 - 7}{4} = \frac{-4}{4} = -1$

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(ii) $5x^2 + 13x + 8 = 0$

Comparing with $ax^2 + bx + c = 0$, we have $a=5$, $b=13$, $c=8$.

Discriminant $\Delta = b^2 - 4ac = (13)^2 - 4(5)(8) = 169 - 160 = 9$.

Since $\Delta = 9 > 0$, the equation has two distinct real roots.

Using the quadratic formula:

$x = \frac{-13 \pm \sqrt{9}}{2(5)}$

$x = \frac{-13 \pm 3}{10}$

The roots are:

$x_1 = \frac{-13 + 3}{10} = \frac{-10}{10} = -1$

$x_2 = \frac{-13 - 3}{10} = \frac{-16}{10} = -\frac{8}{5}$

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(iii) $-3x^2 + 5x + 12 = 0$

Comparing with $ax^2 + bx + c = 0$, we have $a=-3$, $b=5$, $c=12$.

Discriminant $\Delta = b^2 - 4ac = (5)^2 - 4(-3)(12) = 25 + 144 = 169$.

Since $\Delta = 169 > 0$, the equation has two distinct real roots.

Using the quadratic formula:

$x = \frac{-5 \pm \sqrt{169}}{2(-3)}$

$x = \frac{-5 \pm 13}{-6}$

The roots are:

$x_1 = \frac{-5 + 13}{-6} = \frac{8}{-6} = -\frac{4}{3}$

$x_2 = \frac{-5 - 13}{-6} = \frac{-18}{-6} = 3$

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(iv) $-x^2 + 7x – 10 = 0$

Comparing with $ax^2 + bx + c = 0$, we have $a=-1$, $b=7$, $c=-10$.

Discriminant $\Delta = b^2 - 4ac = (7)^2 - 4(-1)(-10) = 49 - 40 = 9$.

Since $\Delta = 9 > 0$, the equation has two distinct real roots.

Using the quadratic formula:

$x = \frac{-7 \pm \sqrt{9}}{2(-1)}$

$x = \frac{-7 \pm 3}{-2}$

The roots are:

$x_1 = \frac{-7 + 3}{-2} = \frac{-4}{-2} = 2$

$x_2 = \frac{-7 - 3}{-2} = \frac{-10}{-2} = 5$

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(v) $x^2 + 2\sqrt{2}x – 6 = 0$

Comparing with $ax^2 + bx + c = 0$, we have $a=1$, $b=2\sqrt{2}$, $c=-6$.

Discriminant $\Delta = b^2 - 4ac = (2\sqrt{2})^2 - 4(1)(-6) = (4 \times 2) + 24 = 8 + 24 = 32$.

Since $\Delta = 32 > 0$, the equation has two distinct real roots.

Using the quadratic formula:

$x = \frac{-(2\sqrt{2}) \pm \sqrt{32}}{2(1)}$

$x = \frac{-2\sqrt{2} \pm \sqrt{16 \times 2}}{2}$

$x = \frac{-2\sqrt{2} \pm 4\sqrt{2}}{2}$

The roots are:

$x_1 = \frac{-2\sqrt{2} + 4\sqrt{2}}{2} = \frac{2\sqrt{2}}{2} = \sqrt{2}$

$x_2 = \frac{-2\sqrt{2} - 4\sqrt{2}}{2} = \frac{-6\sqrt{2}}{2} = -3\sqrt{2}$

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(vi) $x^2 – 3\sqrt{5}x + 10 = 0$

Comparing with $ax^2 + bx + c = 0$, we have $a=1$, $b=-3\sqrt{5}$, $c=10.

Discriminant $\Delta = b^2 - 4ac = (-3\sqrt{5})^2 - 4(1)(10) = (9 \times 5) - 40 = 45 - 40 = 5$.

Since $\Delta = 5 > 0$, the equation has two distinct real roots.

Using the quadratic formula:

$x = \frac{-(-3\sqrt{5}) \pm \sqrt{5}}{2(1)}$

$x = \frac{3\sqrt{5} \pm \sqrt{5}}{2}$

The roots are:

$x_1 = \frac{3\sqrt{5} + \sqrt{5}}{2} = \frac{4\sqrt{5}}{2} = 2\sqrt{5}$

$x_2 = \frac{3\sqrt{5} - \sqrt{5}}{2} = \frac{2\sqrt{5}}{2} = \sqrt{5}$

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(vii) $\frac{1}{2}$x² - $\sqrt{11}$x + 1 = 0

Multiplying the entire equation by 2 to remove the fraction, we get:

$x^2 - 2\sqrt{11}x + 2 = 0$

Comparing with $ax^2 + bx + c = 0$, we have $a=1$, $b=-2\sqrt{11}$, $c=2.

Discriminant $\Delta = b^2 - 4ac = (-2\sqrt{11})^2 - 4(1)(2) = (4 \times 11) - 8 = 44 - 8 = 36$.

Since $\Delta = 36 > 0$, the equation has two distinct real roots.

Using the quadratic formula:

$x = \frac{-(-2\sqrt{11}) \pm \sqrt{36}}{2(1)}$

$x = \frac{2\sqrt{11} \pm 6}{2}$

The roots are:

$x_1 = \frac{2\sqrt{11} + 6}{2} = \frac{2(\sqrt{11} + 3)}{2} = \sqrt{11} + 3$

$x_2 = \frac{2\sqrt{11} - 6}{2} = \frac{2(\sqrt{11} - 3)}{2} = \sqrt{11} - 3$

We will find the roots of the given quadratic equations by the factorisation method.

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(i) $2x^2 + \frac{5}{3} x – 2 = 0$

First, multiply the equation by 3 to eliminate the fraction:

$3 \times (2x^2 + \frac{5}{3} x – 2) = 3 \times 0$

$6x^2 + 5x – 6 = 0$

Here, $a=6$, $b=5$, $c=-6$. We need to find two numbers whose product is $ac = 6 \times (-6) = -36$ and whose sum is $b = 5$. The numbers are $9$ and $-4$ ($9 \times -4 = -36$ and $9 + (-4) = 5$).

Split the middle term $5x$ as $9x - 4x$:

$6x^2 + 9x - 4x - 6 = 0$

Group the terms and factor:

$(6x^2 + 9x) - (4x + 6) = 0$

$3x(2x + 3) - 2(2x + 3) = 0$

Factor out the common binomial term $(2x + 3)$:

$(2x + 3)(3x - 2) = 0$

Set each factor to zero:

$2x + 3 = 0$ or $3x - 2 = 0$

$2x = -3 \implies x = -\frac{3}{2}$

$3x = 2 \implies x = \frac{2}{3}$

The roots are $x = -\frac{3}{2}$ and $x = \frac{2}{3}$.

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(ii) $\frac{2}{5} x^2 - x - \frac{3}{5} = 0$

First, multiply the equation by 5 to eliminate the fractions:

$5 \times (\frac{2}{5} x^2 - x - \frac{3}{5}) = 5 \times 0$

$2x^2 - 5x - 3 = 0$

Here, $a=2$, $b=-5$, $c=-3$. We need to find two numbers whose product is $ac = 2 \times (-3) = -6$ and whose sum is $b = -5$. The numbers are $-6$ and $1$ ($-6 \times 1 = -6$ and $-6 + 1 = -5$).

Split the middle term $-5x$ as $-6x + x$:

$2x^2 - 6x + x - 3 = 0$

Group the terms and factor:

$(2x^2 - 6x) + (x - 3) = 0$

$2x(x - 3) + 1(x - 3) = 0$

Factor out the common binomial term $(x - 3)$:

$(x - 3)(2x + 1) = 0$

Set each factor to zero:

$x - 3 = 0$ or $2x + 1 = 0$

$x = 3$

$2x = -1 \implies x = -\frac{1}{2}$

The roots are $x = 3$ and $x = -\frac{1}{2}$.

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(iii) $3\sqrt{2} x^2 – 5x – \sqrt{2} = 0$

Here, $a=3\sqrt{2}$, $b=-5$, $c=-\sqrt{2}$. We need to find two numbers whose product is $ac = (3\sqrt{2}) \times (-\sqrt{2}) = -3 \times 2 = -6$ and whose sum is $b = -5$. The numbers are $-6$ and $1$ ($-6 \times 1 = -6$ and $-6 + 1 = -5$).

Split the middle term $-5x$ as $-6x + x$:

$3\sqrt{2} x^2 - 6x + x - \sqrt{2} = 0$

Note that $6 = 3 \times 2 = 3 \times \sqrt{2} \times \sqrt{2}$.

Group the terms and factor:

$(3\sqrt{2} x^2 - 6x) + (x - \sqrt{2}) = 0$

$3\sqrt{2}x(x - \sqrt{2}) + 1(x - \sqrt{2}) = 0$

Factor out the common binomial term $(x - \sqrt{2})$:

$(x - \sqrt{2})(3\sqrt{2}x + 1) = 0$

Set each factor to zero:

$x - \sqrt{2} = 0$ or $3\sqrt{2}x + 1 = 0$

$x = \sqrt{2}$

$3\sqrt{2}x = -1 \implies x = -\frac{1}{3\sqrt{2}} = -\frac{\sqrt{2}}{6}$ (by rationalizing the denominator)

The roots are $x = \sqrt{2}$ and $x = -\frac{\sqrt{2}}{6}$.

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(iv) $3x^2 + 5\sqrt{5}x – 10 = 0$

Here, $a=3$, $b=5\sqrt{5}$, $c=-10$. We need to find two numbers whose product is $ac = 3 \times (-10) = -30$ and whose sum is $b = 5\sqrt{5}$. Let the numbers be $p\sqrt{5}$ and $q\sqrt{5}$.

Sum: $p\sqrt{5} + q\sqrt{5} = (p+q)\sqrt{5} = 5\sqrt{5} \implies p+q = 5$.

Product: $(p\sqrt{5})(q\sqrt{5}) = 5pq = -30 \implies pq = -6$.

We need two numbers $p$ and $q$ such that $p+q=5$ and $pq=-6$. The numbers are $6$ and $-1$ ($6 \times -1 = -6$ and $6 + (-1) = 5$).

So the terms are $6\sqrt{5}x$ and $-\sqrt{5}x$.

Split the middle term $5\sqrt{5}x$ as $6\sqrt{5}x - \sqrt{5}x$:

$3x^2 + 6\sqrt{5}x - \sqrt{5}x - 10 = 0$

Note that $6 = 2 \times 3$ and $10 = 2 \times 5 = 2 \times \sqrt{5} \times \sqrt{5}$.

Group the terms and factor:

$(3x^2 + 6\sqrt{5}x) - (\sqrt{5}x + 10) = 0$

$3x(x + 2\sqrt{5}) - \sqrt{5}(x + 2\sqrt{5}) = 0$

Factor out the common binomial term $(x + 2\sqrt{5})$:

$(x + 2\sqrt{5})(3x - \sqrt{5}) = 0$

Set each factor to zero:

$x + 2\sqrt{5} = 0$ or $3x - \sqrt{5} = 0$

$x = -2\sqrt{5}$

$3x = \sqrt{5} \implies x = \frac{\sqrt{5}}{3}$

The roots are $x = -2\sqrt{5}$ and $x = \frac{\sqrt{5}}{3}$.

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(v) $21x^2 – 2x + \frac{1}{21} = 0$

First, multiply the equation by 21 to eliminate the fraction:

$21 \times (21x^2 – 2x + \frac{1}{21}) = 21 \times 0$

$441x^2 - 42x + 1 = 0$

Here, $a=441$, $b=-42$, $c=1$. We need to find two numbers whose product is $ac = 441 \times 1 = 441$ and whose sum is $b = -42$. Note that $441 = 21^2$ and $-42 = -21 - 21$. The numbers are $-21$ and $-21$ ($-21 \times -21 = 441$ and $-21 + (-21) = -42$).

Split the middle term $-42x$ as $-21x - 21x$:

$441x^2 - 21x - 21x + 1 = 0$

Group the terms and factor:

$(441x^2 - 21x) - (21x - 1) = 0$

$21x(21x - 1) - 1(21x - 1) = 0$

Factor out the common binomial term $(21x - 1)$:

$(21x - 1)(21x - 1) = 0$

This is a perfect square trinomial $(21x-1)^2 = 0$.

Set the factor to zero:

$21x - 1 = 0$

$21x = 1$

$x = \frac{1}{21}$

The equation has two equal real roots (or one real root with multiplicity 2).

The roots are $x = \frac{1}{21}$ and $x = \frac{1}{21}$.

Solution:

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The given quadratic equation is $6x^2 – 7x + 2 = 0$.

To check for real roots, we first calculate the discriminant $\Delta = b^2 - 4ac$.

Comparing the given equation with the standard form $ax^2 + bx + c = 0$, we have $a=6$, $b=-7$, and $c=2$.

Discriminant $\Delta = (-7)^2 - 4(6)(2)$

$\Delta = 49 - 48$

$\Delta = 1$

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Since the discriminant $\Delta = 1 > 0$, the equation has two distinct real roots.

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Now, we find the roots using the method of completing the square.

The equation is $6x^2 – 7x + 2 = 0$.

Divide the entire equation by the coefficient of $x^2$, which is 6:

$\frac{6x^2}{6} - \frac{7x}{6} + \frac{2}{6} = \frac{0}{6}$

$x^2 - \frac{7}{6}x + \frac{1}{3} = 0$

Move the constant term to the right side of the equation:

$x^2 - \frac{7}{6}x = -\frac{1}{3}$

Now, we need to add the square of half of the coefficient of $x$ to both sides of the equation. The coefficient of $x$ is $-\frac{7}{6}$. Half of the coefficient is $\frac{1}{2} \times (-\frac{7}{6}) = -\frac{7}{12}$. The square of half the coefficient is $(-\frac{7}{12})^2 = \frac{49}{144}$.

Add $\frac{49}{144}$ to both sides:

$x^2 - \frac{7}{6}x + \frac{49}{144} = -\frac{1}{3} + \frac{49}{144}$

The left side is now a perfect square trinomial $(x - \frac{7}{12})^2$. Simplify the right side:

$-\frac{1}{3} + \frac{49}{144} = -\frac{1 \times 48}{3 \times 48} + \frac{49}{144} = -\frac{48}{144} + \frac{49}{144} = \frac{49 - 48}{144} = \frac{1}{144}$

So the equation becomes:

$(x - \frac{7}{12})^2 = \frac{1}{144}$

Take the square root of both sides:

$x - \frac{7}{12} = \pm \sqrt{\frac{1}{144}}$

$x - \frac{7}{12} = \pm \frac{1}{12}$

Now, solve for $x$ using both the positive and negative values:

Case 1:

$x - \frac{7}{12} = \frac{1}{12}$

$x = \frac{7}{12} + \frac{1}{12}$

$x = \frac{7+1}{12} = \frac{8}{12} = \frac{\cancel{8}^2}{\cancel{12}^3} = \frac{2}{3}$

Case 2:

$x - \frac{7}{12} = -\frac{1}{12}$

$x = \frac{7}{12} - \frac{1}{12}$

$x = \frac{7-1}{12} = \frac{6}{12} = \frac{\cancel{6}^1}{\cancel{12}^2} = \frac{1}{2}$

---

The roots of the quadratic equation $6x^2 – 7x + 2 = 0$ are $\frac{2}{3}$ and $\frac{1}{2}$.

Solution:

---

Let the actual marks scored by Ajita in the mathematics test be $x$.

According to the problem, the maximum marks for the test are 30.

If she had scored 10 more marks, her marks would have been $x + 10$.

The problem states that 9 times these increased marks would have been the square of her actual marks.

So, we can write the equation:

$9(x + 10) = x^2$

---

Now, let's simplify and solve this quadratic equation:

$9x + 90 = x^2$

Rearrange the terms to get the standard quadratic form $ax^2 + bx + c = 0$:

$x^2 - 9x - 90 = 0$

We can solve this by factorisation.

We need two numbers whose product is $ac = 1 \times (-90) = -90$ and whose sum is $b = -9$. The numbers are $-15$ and $6$ (since $-15 \times 6 = -90$ and $-15 + 6 = -9$).

Split the middle term $-9x$ as $-15x + 6x$:

$x^2 - 15x + 6x - 90 = 0$

Group the terms and factor:

$(x^2 - 15x) + (6x - 90) = 0$

$x(x - 15) + 6(x - 15) = 0$

Factor out the common binomial term $(x - 15)$:

$(x - 15)(x + 6) = 0$

Set each factor to zero to find the possible values of $x$:

$x - 15 = 0$ or $x + 6 = 0$

$x = 15$ or $x = -6$

---

Since the marks obtained in a test cannot be negative, $x = -6$ is not a valid solution.

The only valid solution is $x = 15$.

The maximum marks are 30, and $15$ is a valid score within this range.

---

Therefore, the actual marks scored by Ajita in the test are 15.

Solution:

---

Let the original average speed of the train be $v$ km/h.

We know that $\text{Time} = \frac{\text{Distance}}{\text{Speed}}$.

---

For the first part of the journey:

Distance covered = 63 km

Average speed = $v$ km/h

Time taken, $t_1 = \frac{63}{v}$ hours.

---

For the second part of the journey:

Distance covered = 72 km

Average speed = $(v+6)$ km/h (6 km/h more than the original speed)

Time taken, $t_2 = \frac{72}{v+6}$ hours.

---

The total time taken for the entire journey is given as 3 hours.

Total time = $t_1 + t_2 = 3$ hours.

So, we have the equation:

$\frac{63}{v} + \frac{72}{v+6} = 3$

---

To solve this equation for $v$, first find a common denominator, which is $v(v+6)$. Assume $v \neq 0$ and $v+6 \neq 0$. Since speed must be positive, this assumption holds for a valid solution.

Multiply both sides by $v(v+6)$:

$v(v+6) \left( \frac{63}{v} + \frac{72}{v+6} \right) = 3 v(v+6)$

$63(v+6) + 72v = 3v(v+6)$

Expand both sides:

$63v + 378 + 72v = 3v^2 + 18v$

Combine like terms:

$135v + 378 = 3v^2 + 18v$

Rearrange into the standard quadratic form $av^2 + bv + c = 0$:

$3v^2 + 18v - 135v - 378 = 0$

$3v^2 - 117v - 378 = 0$

Divide the entire equation by 3 to simplify:

$\frac{3v^2}{3} - \frac{117v}{3} - \frac{378}{3} = 0$

$v^2 - 39v - 126 = 0$

---

Now, we solve the quadratic equation $v^2 - 39v - 126 = 0$ by factorisation.

We need to find two numbers whose product is $1 \times (-126) = -126$ and whose sum is $-39$. The numbers are $-42$ and $3$. (Since $-42 \times 3 = -126$ and $-42 + 3 = -39$).

Split the middle term $-39v$ as $-42v + 3v$:

$v^2 - 42v + 3v - 126 = 0$

Group the terms and factor common factors:

$(v^2 - 42v) + (3v - 126) = 0$

$v(v - 42) + 3(v - 42) = 0$

Factor out the common binomial term $(v - 42)$:

$(v - 42)(v + 3) = 0$

---

Set each factor equal to zero to find the possible values of $v$:

$v - 42 = 0$ or $v + 3 = 0$

If $v - 42 = 0$, then $v = 42$.

If $v + 3 = 0$, then $v = -3$.

---

Since speed is a physical quantity, it cannot be negative. Therefore, $v = -3$ km/h is not a valid solution for the original average speed.

The valid solution is $v = 42$ km/h.

---

Thus, the original average speed of the train is 42 km/h.

To check for real roots of a quadratic equation $ax^2 + bx + c = 0$, we calculate the discriminant $\Delta = b^2 - 4ac$.

If $\Delta \ge 0$, real roots exist. If $\Delta < 0$, real roots do not exist.

If real roots exist, they can be found using the quadratic formula: $x = \frac{-b \pm \sqrt{\Delta}}{2a}$.

---

(i) $8x^2 + 2x – 3 = 0$

Here, $a=8$, $b=2$, $c=-3$.

Discriminant $\Delta = b^2 - 4ac = (2)^2 - 4(8)(-3) = 4 + 96 = 100$.

Since $\Delta = 100 > 0$, the equation has two distinct real roots.

Using the quadratic formula:

$x = \frac{-2 \pm \sqrt{100}}{2(8)}$

$x = \frac{-2 \pm 10}{16}$

The roots are:

$x_1 = \frac{-2 + 10}{16} = \frac{8}{16} = \frac{1}{2}$

$x_2 = \frac{-2 - 10}{16} = \frac{-12}{16} = -\frac{3}{4}$

The real roots are $\frac{1}{2}$ and $-\frac{3}{4}$.

---

(ii) $-2x^2 + 3x + 2 = 0$

Here, $a=-2$, $b=3$, $c=2$.

Discriminant $\Delta = b^2 - 4ac = (3)^2 - 4(-2)(2) = 9 + 16 = 25$.

Since $\Delta = 25 > 0$, the equation has two distinct real roots.

Using the quadratic formula:

$x = \frac{-3 \pm \sqrt{25}}{2(-2)}$

$x = \frac{-3 \pm 5}{-4}$

The roots are:

$x_1 = \frac{-3 + 5}{-4} = \frac{2}{-4} = -\frac{1}{2}$

$x_2 = \frac{-3 - 5}{-4} = \frac{-8}{-4} = 2$

The real roots are $-\frac{1}{2}$ and $2$.

---

(iii) $5x^2 – 2x – 10 = 0$

Here, $a=5$, $b=-2$, $c=-10$.

Discriminant $\Delta = b^2 - 4ac = (-2)^2 - 4(5)(-10) = 4 + 200 = 204$.

Since $\Delta = 204 > 0$, the equation has two distinct real roots.

Using the quadratic formula:

$x = \frac{-(-2) \pm \sqrt{204}}{2(5)}$

$x = \frac{2 \pm \sqrt{4 \times 51}}{10}$

$x = \frac{2 \pm 2\sqrt{51}}{10}$

$x = \frac{2(1 \pm \sqrt{51})}{10}$

$x = \frac{1 \pm \sqrt{51}}{5}$

The real roots are $\frac{1 + \sqrt{51}}{5}$ and $\frac{1 - \sqrt{51}}{5}$.

---

(iv) $\frac{1}{2x \;-\; 3} + \frac{1}{x \;-\; 5}$ = 1

Given $x \neq \frac{3}{2}$ and $x \neq 5$.

Find a common denominator for the left side:

$\frac{(x - 5) + (2x - 3)}{(2x - 3)(x - 5)} = 1$

Simplify the numerator and the denominator:

$\frac{x - 5 + 2x - 3}{2x^2 - 10x - 3x + 15} = 1$

$\frac{3x - 8}{2x^2 - 13x + 15} = 1$

Multiply both sides by the denominator $(2x^2 - 13x + 15)$ (which is non-zero as $x \neq \frac{3}{2}, 5$):

$3x - 8 = 1 \times (2x^2 - 13x + 15)$

$3x - 8 = 2x^2 - 13x + 15$

Rearrange into the standard quadratic form $ax^2 + bx + c = 0$:

$0 = 2x^2 - 13x - 3x + 15 + 8$

$2x^2 - 16x + 23 = 0$

Here, $a=2$, $b=-16$, $c=23$.

Discriminant $\Delta = b^2 - 4ac = (-16)^2 - 4(2)(23) = 256 - 184 = 72$.

Since $\Delta = 72 > 0$, the equation has two distinct real roots.

Using the quadratic formula:

$x = \frac{-(-16) \pm \sqrt{72}}{2(2)}$

$x = \frac{16 \pm \sqrt{36 \times 2}}{4}$

$x = \frac{16 \pm 6\sqrt{2}}{4}$

$x = \frac{2(8 \pm 3\sqrt{2})}{4}$

$x = \frac{8 \pm 3\sqrt{2}}{2}$

The real roots are $\frac{8 + 3\sqrt{2}}{2}$ and $\frac{8 - 3\sqrt{2}}{2}$.

---

(v) $x^2 + 5\sqrt{5}$x – 70 = 0

Here, $a=1$, $b=5\sqrt{5}$, $c=-70$.

Discriminant $\Delta = b^2 - 4ac = (5\sqrt{5})^2 - 4(1)(-70) = (25 \times 5) + 280 = 125 + 280 = 405$.

Since $\Delta = 405 > 0$, the equation has two distinct real roots.

Using the quadratic formula:

$x = \frac{-(5\sqrt{5}) \pm \sqrt{405}}{2(1)}$

$x = \frac{-5\sqrt{5} \pm \sqrt{81 \times 5}}{2}$

$x = \frac{-5\sqrt{5} \pm 9\sqrt{5}}{2}$

The roots are:

$x_1 = \frac{-5\sqrt{5} + 9\sqrt{5}}{2} = \frac{4\sqrt{5}}{2} = 2\sqrt{5}$

$x_2 = \frac{-5\sqrt{5} - 9\sqrt{5}}{2} = \frac{-14\sqrt{5}}{2} = -7\sqrt{5}$

The real roots are $2\sqrt{5}$ and $-7\sqrt{5}$.

Solution:

---

Let the required natural number be $x$.

According to the problem statement, the square of the number diminished by 84 is $x^2 - 84$.

Thrice of 8 more than the number is $3 \times (x + 8)$.

The problem states that these two expressions are equal.

So, we can write the equation:

$x^2 - 84 = 3(x + 8)$

---

Now, let's simplify and solve the equation:

Expand the right side:

$x^2 - 84 = 3x + 24$

Rearrange the terms to form a quadratic equation in standard form $ax^2 + bx + c = 0$:

$x^2 - 3x - 84 - 24 = 0$

$x^2 - 3x - 108 = 0$

---

We can solve this quadratic equation by factorisation.

We need to find two numbers whose product is $ac = 1 \times (-108) = -108$ and whose sum is $b = -3$. The numbers are $-12$ and $9$, since $(-12) \times 9 = -108$ and $-12 + 9 = -3$.

Split the middle term $-3x$ as $-12x + 9x$:

$x^2 - 12x + 9x - 108 = 0$

Group the terms and factor common factors:

$(x^2 - 12x) + (9x - 108) = 0$

$x(x - 12) + 9(x - 12) = 0$

Factor out the common binomial term $(x - 12)$:

$(x - 12)(x + 9) = 0$

---

Set each factor to zero to find the possible values of $x$:

$x - 12 = 0$ or $x + 9 = 0$

$x = 12$ or $x = -9$

---

The problem asks for a natural number. Natural numbers are positive integers (1, 2, 3, ...).

The possible solutions are $x = 12$ and $x = -9$.

Since $-9$ is not a natural number, we discard this solution.

The natural number satisfying the condition is $x = 12$.

---

Let's verify the answer: Square of 12 diminished by 84 is $12^2 - 84 = 144 - 84 = 60$. Thrice of 8 more than 12 is $3 \times (12 + 8) = 3 \times 20 = 60$. The values are equal, so the answer is correct.

The required natural number is 12.

Solution:

---

Let the required natural number be $x$.

According to the problem statement:

"A natural number, when increased by 12" is $x + 12$.

"160 times its reciprocal" is $160 \times \frac{1}{x} = \frac{160}{x}$.

The problem states that these two quantities are equal:

$x + 12 = \frac{160}{x}$

---

To solve this equation, multiply both sides by $x$. Since $x$ is a natural number, $x \neq 0$.

$x(x + 12) = x \left( \frac{160}{x} \right)$

$x^2 + 12x = 160$

Rearrange the terms to form a standard quadratic equation $ax^2 + bx + c = 0$:

$x^2 + 12x - 160 = 0$

---

We can solve this quadratic equation by factorisation.

We need to find two numbers whose product is $ac = 1 \times (-160) = -160$ and whose sum is $b = 12$. The numbers are $20$ and $-8$, since $20 \times (-8) = -160$ and $20 + (-8) = 12$.

Split the middle term $12x$ as $20x - 8x$:

$x^2 + 20x - 8x - 160 = 0$

Group the terms and factor common factors:

$(x^2 + 20x) - (8x + 160) = 0$

$x(x + 20) - 8(x + 20) = 0$

Factor out the common binomial term $(x + 20)$:

$(x + 20)(x - 8) = 0$

---

Set each factor to zero to find the possible values of $x$:

$x + 20 = 0$ or $x - 8 = 0$

$x = -20$ or $x = 8$

---

The problem asks for a natural number. Natural numbers are positive integers (1, 2, 3, ...).

The possible solutions are $x = -20$ and $x = 8$.

Since $-20$ is not a natural number, we discard this solution.

The natural number satisfying the condition is $x = 8$.

---

Let's verify the answer: The number increased by 12 is $8 + 12 = 20$. 160 times its reciprocal is $160 \times \frac{1}{8} = \frac{160}{8} = 20$. The values are equal, so the answer is correct.

The required natural number is 8.

Solution:

---

Let the original uniform speed of the train be $v$ km/h.

The distance to be travelled is 360 km.

The time taken to travel this distance at the original speed is given by:

Original time $t_1 = \frac{\text{Distance}}{\text{Speed}} = \frac{360}{v}$ hours.

---

If the speed were 5 km/h more, the new speed would be $(v+5)$ km/h.

The time taken to travel the same distance (360 km) at this increased speed is:

New time $t_2 = \frac{360}{v+5}$ hours.

---

According to the problem, the train would have taken 48 minutes less to travel the same distance with the increased speed.

First, convert 48 minutes to hours: $48 \text{ minutes} = \frac{48}{60} \text{ hours} = \frac{4}{5} \text{ hours}$.

The difference between the original time and the new time is $\frac{4}{5}$ hours:

$t_1 - t_2 = \frac{4}{5}$

Substitute the expressions for $t_1$ and $t_2$:

$\frac{360}{v} - \frac{360}{v+5} = \frac{4}{5}$

---

Now, we solve this equation for $v$. Find a common denominator on the left side, which is $v(v+5)$. Assume $v > 0$ since it's a speed.

$\frac{360(v+5) - 360v}{v(v+5)} = \frac{4}{5}$

Simplify the numerator:

$\frac{360v + 1800 - 360v}{v^2 + 5v} = \frac{4}{5}$

$\frac{1800}{v^2 + 5v} = \frac{4}{5}$

Cross-multiply:

$5 \times 1800 = 4 \times (v^2 + 5v)$

$9000 = 4v^2 + 20v$

Rearrange into the standard quadratic form $av^2 + bv + c = 0$:

$4v^2 + 20v - 9000 = 0$

Divide the entire equation by 4:

$\frac{4v^2}{4} + \frac{20v}{4} - \frac{9000}{4} = \frac{0}{4}$

$v^2 + 5v - 2250 = 0$

---

We can solve this quadratic equation by factorisation. We need two numbers whose product is $1 \times (-2250) = -2250$ and whose sum is $5$. The numbers are $50$ and $-45$ ($50 \times -45 = -2250$ and $50 + (-45) = 5$).

Split the middle term $5v$ as $50v - 45v$:

$v^2 + 50v - 45v - 2250 = 0$

Group the terms and factor:

$(v^2 + 50v) - (45v + 2250) = 0$

$v(v + 50) - 45(v + 50) = 0$

Factor out the common binomial term $(v + 50)$:

$(v + 50)(v - 45) = 0$

---

Set each factor to zero to find the possible values of $v$:

$v + 50 = 0$ or $v - 45 = 0$

$v = -50$ or $v = 45$

---

Since the speed of the train cannot be negative, the value $v = -50$ km/h is not a valid solution.

The valid solution for the original speed is $v = 45$ km/h.

---

The original speed of the train is 45 km/h.

Solution:

---

Let Zeba's present or actual age be $x$ years.

If Zeba were younger by 5 years, her age would be $x - 5$ years.

According to the problem, the square of this hypothetical age is 11 more than five times her actual age.

So, we can form the equation:

$(x - 5)^2 = 5x + 11$

---

Now, we solve this equation for $x$.

Expand the left side of the equation $(x - 5)^2 = x^2 - 2(x)(5) + 5^2 = x^2 - 10x + 25$.

The equation becomes:

$x^2 - 10x + 25 = 5x + 11$

Rearrange the terms to get a standard quadratic equation $ax^2 + bx + c = 0$:

$x^2 - 10x - 5x + 25 - 11 = 0$

$x^2 - 15x + 14 = 0$

---

We can solve this quadratic equation by factorisation. We need to find two numbers whose product is $ac = 1 \times 14 = 14$ and whose sum is $b = -15$. The numbers are $-14$ and $-1$, since $(-14) \times (-1) = 14$ and $(-14) + (-1) = -15$.

Split the middle term $-15x$ as $-14x - x$:

$x^2 - 14x - x + 14 = 0$

Group the terms and factor common factors:

$(x^2 - 14x) - (x - 14) = 0$

$x(x - 14) - 1(x - 14) = 0$

Factor out the common binomial term $(x - 14)$:

$(x - 14)(x - 1) = 0$

---

Set each factor to zero to find the possible values of $x$:

$x - 14 = 0$ or $x - 1 = 0$

$x = 14$ or $x = 1$

---

Since $x$ represents Zeba's age, it must be a positive value. Both 14 and 1 are positive. However, the problem mentions "If Zeba were younger by 5 years". This implies that her actual age must be at least 5 years so that $x-5$ is a non-negative age.

If $x=1$, then $x-5 = 1-5 = -4$, which is not a valid age.

If $x=14$, then $x-5 = 14-5 = 9$, which is a valid age.

Therefore, the only valid solution is $x = 14$.

---

Zeba's actual age is 14 years.

Let's verify: If her age were 9 (14-5), the square would be $9^2 = 81$. Five times her actual age (14) is $5 \times 14 = 70$. 11 more than five times her actual age is $70 + 11 = 81$. The values match.

Zeba's age now is 14 years.

Solution:

---

Let Nisha's present age be $n$ years.

Let Asha's present age be $a$ years.

---

According to the first condition: "At present Asha’s age (in years) is 2 more than the square of her daughter Nisha’s age."

$a = n^2 + 2$

---

According to the second condition: "When Nisha grows to her mother’s present age..."

Nisha's present age is $n$. Nisha grows to her mother's present age, which is $a$.

The time taken for Nisha to reach age $a$ is $(a - n)$ years.

In $(a - n)$ years, Asha's age will increase by $(a - n)$ years.

Asha's age when Nisha is $a$ years old will be $a + (a - n) = 2a - n$ years.

---

The second condition also states: "...Asha’s age would be one year less than 10 times the present age of Nisha."

So, Asha's age in the future scenario ($2a - n$) is equal to $(10 \times \text{Nisha's present age}) - 1$.

$2a - n = 10n - 1$

$2a = 11n - 1$

---

Now we have a system of two equations:

(1) $a = n^2 + 2$

(2) $2a = 11n - 1$

Substitute the expression for $a$ from equation (1) into equation (2):

$2(n^2 + 2) = 11n - 1$

Expand and simplify the equation:

$2n^2 + 4 = 11n - 1$

Rearrange the terms to form a standard quadratic equation:

$2n^2 - 11n + 4 + 1 = 0$

$2n^2 - 11n + 5 = 0$

---

We solve this quadratic equation for $n$ by factorisation.

We look for two numbers whose product is $ac = 2 \times 5 = 10$ and whose sum is $b = -11$. These numbers are $-10$ and $-1$.

Split the middle term $-11n$ as $-10n - n$:

$2n^2 - 10n - n + 5 = 0$

Group the terms and factor common factors:

$(2n^2 - 10n) - (n - 5) = 0$

$2n(n - 5) - 1(n - 5) = 0$

Factor out the common binomial term $(n - 5)$:

$(n - 5)(2n - 1) = 0$

---

Set each factor to zero to find the possible values for $n$:

$n - 5 = 0$ or $2n - 1 = 0$

$n = 5$ or $2n = 1 \implies n = \frac{1}{2}$

---

Since $n$ represents Nisha's age, it must be a positive value. Both $n=5$ and $n=\frac{1}{2}$ are positive.

Let's find the corresponding ages for Asha using $a = n^2 + 2$ and check if the future condition holds.

Case 1: If $n = 5$ (Nisha's age is 5 years)

$a = (5)^2 + 2 = 25 + 2 = 27$ (Asha's age is 27 years).

Check the second condition: When Nisha grows to 27 years, the time elapsed is $27 - 5 = 22$ years. Asha's age will be $27 + 22 = 49$ years. Ten times Nisha's present age is $10 \times 5 = 50$. One year less than 10 times Nisha's present age is $50 - 1 = 49$. This matches Asha's age (49 years). So, $n=5$ is a valid solution.

Case 2: If $n = \frac{1}{2}$ (Nisha's age is 0.5 years)

$a = (\frac{1}{2})^2 + 2 = \frac{1}{4} + 2 = \frac{1}{4} + \frac{8}{4} = \frac{9}{4} = 2.25$ (Asha's age is 2.25 years).

Check the second condition: When Nisha grows to 2.25 years, the time elapsed is $2.25 - 0.5 = 1.75$ years. Asha's age will be $2.25 + 1.75 = 4$ years. Ten times Nisha's present age is $10 \times \frac{1}{2} = 5$. One year less than 10 times Nisha's present age is $5 - 1 = 4$. This matches Asha's age (4 years). So, $n=\frac{1}{2}$ also satisfies the mathematical conditions.

However, in real-world age problems, we usually consider whole numbers for age unless specified otherwise, especially when talking about "growing to a certain age". Also, an age of 0.5 years (6 months) might be unusual contextually in such a problem. The structure of the problem suggests a more typical age scenario.

Considering typical age progression and the phrasing "When Nisha grows to her mother’s present age", it is more likely that Nisha is old enough to 'grow' in a significant sense, and whole number ages are expected unless the problem implies fractional ages (e.g., talking about infants). The context of "square of age" and "10 times the age" points towards integer ages for simplicity and typical problem settings.

Therefore, we consider the natural number solution for Nisha's age.

---

The valid ages are Nisha's age = 5 years and Asha's age = 27 years.

The present age of Nisha is 5 years and the present age of Asha is 27 years.

Solution:

---

Let the dimensions of the rectangular lawn be: Length = 50 m, Breadth = 40 m.

Area of the rectangular lawn = Length $\times$ Breadth = $50 \times 40 = 2000$ m$^2$.

---

A rectangular pond is to be constructed in the centre of the lawn, with grass surrounding it. Let the uniform width of the grass surrounding the pond be $x$ metres.

The dimensions of the rectangular pond will be:

Length of pond = (Length of lawn) - $2 \times$ (Width of grass) = $50 - 2x$ metres.

Breadth of pond = (Breadth of lawn) - $2 \times$ (Width of grass) = $40 - 2x$ metres.

For the pond to have valid dimensions, its length and breadth must be positive. Thus, $50 - 2x > 0 \implies 2x < 50 \implies x < 25$, and $40 - 2x > 0 \implies 2x < 40 \implies x < 20$. Since $x$ is a width, $x \ge 0$. Combining these, we have $0 \le x < 20$.

---

Area of the rectangular pond = (Length of pond) $\times$ (Breadth of pond) = $(50 - 2x)(40 - 2x)$ m$^2$.

---

The area of the grass surrounding the pond is the difference between the area of the lawn and the area of the pond.

Area of grass = Area of lawn - Area of pond

We are given that the area of the grass surrounding the pond is 1184 m$^2$.

$1184 = 2000 - (50 - 2x)(40 - 2x)$

---

Rearrange the equation to solve for $x$:

$(50 - 2x)(40 - 2x) = 2000 - 1184$

$(50 - 2x)(40 - 2x) = 816$

Expand the left side of the equation:

$50 \times 40 + 50 \times (-2x) + (-2x) \times 40 + (-2x) \times (-2x) = 816$

$2000 - 100x - 80x + 4x^2 = 816$

$4x^2 - 180x + 2000 = 816$

Move all terms to one side to form a quadratic equation:

$4x^2 - 180x + 2000 - 816 = 0$

$4x^2 - 180x + 1184 = 0$

Divide the equation by 4 to simplify:

$\frac{4x^2}{4} - \frac{180x}{4} + \frac{1184}{4} = \frac{0}{4}$

$x^2 - 45x + 296 = 0$

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We solve this quadratic equation for $x$ by factorisation.

We need to find two numbers whose product is $1 \times 296 = 296$ and whose sum is $-45$. By considering the factors of 296, we find that $-8$ and $-37$ satisfy these conditions ($-8 \times -37 = 296$ and $-8 + (-37) = -45$).

Split the middle term $-45x$ as $-8x - 37x$:

$x^2 - 8x - 37x + 296 = 0$

Group the terms and factor common factors:

$(x^2 - 8x) - (37x - 296) = 0$

$x(x - 8) - 37(x - 8) = 0$

Factor out the common binomial term $(x - 8)$:

$(x - 8)(x - 37) = 0$

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Set each factor to zero to find the possible values of $x$:

$x - 8 = 0 \implies x = 8$

$x - 37 = 0 \implies x = 37$

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We found two possible values for the width of the grass, $x=8$ m and $x=37$ m.

However, we established the condition that $0 \le x < 20$ for the pond dimensions to be positive.

The value $x=37$ does not satisfy the condition $x < 20$. If $x=37$, the pond dimensions would be $50 - 2(37) = -24$ m and $40 - 2(37) = -34$ m, which are not physically possible.

The value $x=8$ satisfies the condition $0 \le 8 < 20$. So, the width of the grass surrounding the pond is 8 m.

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Now, we find the dimensions of the pond using $x=8$ m:

Length of the pond = $50 - 2x = 50 - 2(8) = 50 - 16 = 34$ metres.

Breadth of the pond = $40 - 2x = 40 - 2(8) = 40 - 16 = 24$ metres.

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The length and breadth of the pond are 34 m and 24 m respectively.

Solution:

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Let the time be $t$ minutes past 2 pm.

The time needed by the minutes hand to show 3 pm (i.e., to reach 3:00 pm) from $t$ minutes past 2 pm is the remaining time until the hour mark changes from 2 to 3.

Total minutes from 2 pm to 3 pm is 60 minutes.

So, the time needed to reach 3 pm from $t$ minutes past 2 pm is $(60 - t)$ minutes.

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According to the problem statement, this time is 3 minutes less than $\frac{t^2}{4}$ minutes.

So, we can set up the equation:

$60 - t = \frac{t^2}{4} - 3$

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Now, we need to solve this equation for $t$.

To eliminate the fraction, multiply the entire equation by 4:

$4 \times (60 - t) = 4 \times (\frac{t^2}{4} - 3)$

$240 - 4t = t^2 - 12$

Rearrange the terms to form a standard quadratic equation $at^2 + bt + c = 0$:

$0 = t^2 + 4t - 12 - 240$

$t^2 + 4t - 252 = 0$

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We solve the quadratic equation $t^2 + 4t - 252 = 0$ by factorisation.

We need to find two numbers whose product is $1 \times (-252) = -252$ and whose sum is $4$. The numbers are $18$ and $-14$, since $18 \times (-14) = -252$ and $18 + (-14) = 4$.

Split the middle term $4t$ as $18t - 14t$:

$t^2 + 18t - 14t - 252 = 0$

Group the terms and factor common factors:

$(t^2 + 18t) - (14t + 252) = 0$

$t(t + 18) - 14(t + 18) = 0$

Factor out the common binomial term $(t + 18)$:

$(t + 18)(t - 14) = 0$

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Set each factor to zero to find the possible values of $t$:

$t + 18 = 0$ or $t - 14 = 0$

$t = -18$ or $t = 14$

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$t$ represents the number of minutes past 2 pm. Therefore, $t$ must be a non-negative value and less than 60 (since it's time past 2 pm, not past 3 pm, and before 3 pm). So, $0 \le t < 60$.

The value $t = -18$ is not valid because time cannot be negative in this context.

The value $t = 14$ is valid because $0 \le 14 < 60$.

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The required value of $t$ is 14.

At 14 minutes past 2 pm, the time needed to show 3 pm is $60 - 14 = 46$ minutes. The given expression for this time is $\frac{t^2}{4} - 3 = \frac{14^2}{4} - 3 = \frac{196}{4} - 3 = 49 - 3 = 46$ minutes. This matches.

The value of t is 14.