Chapter 2 Polynomials (Class 10 - Maths NCERT Textbook Solutions)

The number of zeroes of a polynomial $p(x)$ is equal to the number of times the graph of $y = p(x)$ intersects the x-axis.

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(i) In the first graph, the graph intersects the x-axis at one point.

Therefore, the number of zeroes of $p(x)$ is 1.

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(ii) In the second graph, the graph intersects the x-axis at two points.

Therefore, the number of zeroes of $p(x)$ is 2.

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(iii) In the third graph, the graph intersects the x-axis at three points.

Therefore, the number of zeroes of $p(x)$ is 3.

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(iv) In the fourth graph, the graph intersects the x-axis at one point.

Therefore, the number of zeroes of $p(x)$ is 1.

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(v) In the fifth graph, the graph intersects the x-axis at one point.

Therefore, the number of zeroes of $p(x)$ is 1.

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(vi) In the sixth graph, the graph intersects the x-axis at four points.

Therefore, the number of zeroes of $p(x)$ is 4.

The number of zeroes of a polynomial $p(x)$ is equal to the number of times the graph of $y = p(x)$ intersects the x-axis.

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(i) In the first graph, the graph does not intersect the x-axis at any point.

Therefore, the number of zeroes of $p(x)$ is 0.

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(ii) In the second graph, the graph intersects the x-axis at one distinct points.

Therefore, the number of zeroes of $p(x)$ is 1.

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(iii) In the third graph, the graph intersects the x-axis at three distinct points.

Therefore, the number of zeroes of $p(x)$ is 3.

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(iv) In the fourth graph, the graph intersects the x-axis at two point.

Therefore, the number of zeroes of $p(x)$ is 2.

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(v) In the fifth graph, the graph intersects the x-axis at four point.

Therefore, the number of zeroes of $p(x)$ is 4.

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(vi) In the sixth graph, the graph intersects the x-axis at three distinct points.

Therefore, the number of zeroes of $p(x)$ is 3.

Given:

The quadratic polynomial is $p(x) = x^2 + 7x + 10$.

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To Find:

The zeroes of the polynomial and verify the relationship between the zeroes and the coefficients.

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Solution:

To find the zeroes of the polynomial, we need to find the values of $x$ for which $p(x) = 0$.

$x^2 + 7x + 10 = 0$

We can factor the quadratic expression by splitting the middle term. We look for two numbers that multiply to 10 and add up to 7. These numbers are 2 and 5.

$x^2 + 2x + 5x + 10 = 0$

Group the terms and factor:

$(x^2 + 2x) + (5x + 10) = 0$

$x(x + 2) + 5(x + 2) = 0$

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

$(x + 2)(x + 5) = 0$

To find the zeroes, set each factor equal to zero:

$x + 2 = 0 \quad$ or $\quad x + 5 = 0$

$x = -2 \quad$ or $\quad x = -5$

The zeroes of the polynomial $x^2 + 7x + 10$ are $-2$ and $-5$.

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Verification of the relationship between zeroes and coefficients:

Let the zeroes be $\alpha = -2$ and $\beta = -5$.

The given quadratic polynomial is $x^2 + 7x + 10$. Comparing this with the standard form $ax^2 + bx + c$, we have:

$a = 1$

$b = 7$

$c = 10$

Relationship between zeroes and coefficients:

1. Sum of zeroes: $\alpha + \beta = \frac{-b}{a}$

Left Hand Side (LHS): $\alpha + \beta = (-2) + (-5) = -7$

Right Hand Side (RHS): $\frac{-b}{a} = \frac{-(7)}{1} = -7$

The sum of zeroes is verified.

2. Product of zeroes: $\alpha \times \beta = \frac{c}{a}$

Left Hand Side (LHS): $\alpha \times \beta = (-2) \times (-5) = 10$

Right Hand Side (RHS): $\frac{c}{a} = \frac{10}{1} = 10$

The product of zeroes is verified.

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Since both relationships are satisfied, the verification is complete.

Given:

The quadratic polynomial is $p(x) = x^2 - 3$.

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To Find:

The zeroes of the polynomial and verify the relationship between the zeroes and the coefficients.

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Solution:

To find the zeroes of the polynomial, we need to find the values of $x$ for which $p(x) = 0$.

$x^2 - 3 = 0$

Add 3 to both sides:

$x^2 = 3$

Take the square root of both sides:

$x = \pm\sqrt{3}$

The zeroes of the polynomial $x^2 - 3$ are $\sqrt{3}$ and $-\sqrt{3}$.

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Verification of the relationship between zeroes and coefficients:

Let the zeroes be $\alpha = \sqrt{3}$ and $\beta = -\sqrt{3}$.

The given quadratic polynomial is $x^2 - 3$. We can write it in the standard form $ax^2 + bx + c$ as $x^2 + 0x - 3$. Comparing, we have:

$a = 1$

$b = 0$

$c = -3$

Relationship between zeroes and coefficients:

1. Sum of zeroes: $\alpha + \beta = \frac{-b}{a}$

Left Hand Side (LHS): $\alpha + \beta = \sqrt{3} + (-\sqrt{3}) = 0$

Right Hand Side (RHS): $\frac{-b}{a} = \frac{-(0)}{1} = 0$

The sum of zeroes is verified.

2. Product of zeroes: $\alpha \times \beta = \frac{c}{a}$

Left Hand Side (LHS): $\alpha \times \beta = (\sqrt{3}) \times (-\sqrt{3}) = -(\sqrt{3})^2 = -3$

Right Hand Side (RHS): $\frac{c}{a} = \frac{(-3)}{1} = -3$

The product of zeroes is verified.

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Since both relationships are satisfied, the verification is complete.

Given:

Sum of the zeroes = $-3$

Product of the zeroes = $2$

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To Find:

A quadratic polynomial whose sum and product of zeroes are given.

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Solution:

A quadratic polynomial with sum of zeroes $\alpha + \beta$ and product of zeroes $\alpha \times \beta$ can be written in the form:

$p(x) = k(x^2 - (\text{Sum of zeroes})x + (\text{Product of zeroes}))$

where $k$ is any non-zero real number.

Substitute the given values for the sum and product of zeroes:

$p(x) = k(x^2 - (-3)x + (2))$

Simplify the expression:

$p(x) = k(x^2 + 3x + 2)$

For simplicity, we can take $k = 1$.

So, a quadratic polynomial is $x^2 + 3x + 2$.

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Answer:

A quadratic polynomial is $x^2 + 3x + 2$.

Given:

The cubic polynomial $p(x) = 3x^3 - 5x^2 - 11x - 3$.

The given potential zeroes are $3$, $-1$, and $-\frac{1}{3}$.

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To Verify:

1. That $3$, $-1$, and $-\frac{1}{3}$ are the zeroes of $p(x)$.

2. The relationship between the zeroes and the coefficients of $p(x)$.

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Solution:

Verification of Zeroes:

A number $k$ is a zero of a polynomial $p(x)$ if $p(k) = 0$. We substitute the given values into the polynomial $p(x)$.

For $x = 3$:

$p(3) = 3(3)^3 - 5(3)^2 - 11(3) - 3$

$p(3) = 3(27) - 5(9) - 33 - 3$

$p(3) = 81 - 45 - 33 - 3$

$p(3) = 81 - (45 + 33 + 3)$

$p(3) = 81 - 81$

Since $p(3)=0$, $3$ is a zero of the polynomial $p(x)$.

For $x = -1$:

$p(-1) = 3(-1)^3 - 5(-1)^2 - 11(-1) - 3$

$p(-1) = 3(-1) - 5(1) - (-11) - 3$

$p(-1) = -3 - 5 + 11 - 3$

$p(-1) = (-3 - 5 - 3) + 11$

$p(-1) = -11 + 11$

Since $p(-1)=0$, $-1$ is a zero of the polynomial $p(x)$.

For $x = -\frac{1}{3}$:

$p(-\frac{1}{3}) = 3(-\frac{1}{3})^3 - 5(-\frac{1}{3})^2 - 11(-\frac{1}{3}) - 3$

$p(-\frac{1}{3}) = 3(-\frac{1}{27}) - 5(\frac{1}{9}) + \frac{11}{3} - 3$

$p(-\frac{1}{3}) = -\frac{3}{27} - \frac{5}{9} + \frac{11}{3} - 3$

$p(-\frac{1}{3}) = -\frac{1}{9} - \frac{5}{9} + \frac{11}{3} - 3$

$p(-\frac{1}{3}) = \frac{-1}{9} - \frac{5}{9} + \frac{11 \times 3}{3 \times 3} - \frac{3 \times 9}{1 \times 9}$

$p(-\frac{1}{3}) = \frac{-1 - 5}{9} + \frac{33}{9} - \frac{27}{9}$

$p(-\frac{1}{3}) = \frac{-6 + 33 - 27}{9}$

$p(-\frac{1}{3}) = \frac{27 - 27}{9} = \frac{0}{9}$

Since $p(-\frac{1}{3})=0$, $-\frac{1}{3}$ is a zero of the polynomial $p(x)$.

Thus, we have verified that $3$, $-1$, and $-\frac{1}{3}$ are the zeroes of the cubic polynomial $p(x) = 3x^3 - 5x^2 - 11x - 3$.

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Verification of the relationship between zeroes and coefficients:

Let the zeroes be $\alpha = 3$, $\beta = -1$, and $\gamma = -\frac{1}{3}$.

The given cubic polynomial is $p(x) = 3x^3 - 5x^2 - 11x - 3$.

Comparing this polynomial with the standard form of a cubic polynomial, $ax^3 + bx^2 + cx + d$, we identify the coefficients:

The relationships between the zeroes ($\alpha, \beta, \gamma$) and the coefficients ($a, b, c, d$) of a cubic polynomial are given by:

1. Sum of zeroes: $\alpha + \beta + \gamma = -\frac{b}{a}$

2. Sum of the products of zeroes taken two at a time: $\alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a}$

3. Product of zeroes: $\alpha\beta\gamma = -\frac{d}{a}$

Let's verify these relationships using the given zeroes and the coefficients we found.

Relationship 1: Sum of zeroes

LHS: $\alpha + \beta + \gamma = 3 + (-1) + (-\frac{1}{3})$

$\alpha + \beta + \gamma = 3 - 1 - \frac{1}{3}$

$\alpha + \beta + \gamma = 2 - \frac{1}{3}$

RHS: $-\frac{b}{a}$

Using values from (5) and (4):

From (8) and (9), we see that LHS = RHS. The first relationship is verified.

Relationship 2: Sum of the products of zeroes taken two at a time

LHS: $\alpha\beta + \beta\gamma + \gamma\alpha = (3)(-1) + (-1)(-\frac{1}{3}) + (-\frac{1}{3})(3)$

$\alpha\beta + \beta\gamma + \gamma\alpha = -3 + \frac{1}{3} - \frac{3}{3}$

$\alpha\beta + \beta\gamma + \gamma\alpha = -3 + \frac{1}{3} - 1$

$\alpha\beta + \beta\gamma + \gamma\alpha = -4 + \frac{1}{3}$

RHS: $\frac{c}{a}$

Using values from (6) and (4):

From (10) and (11), we see that LHS = RHS. The second relationship is verified.

Relationship 3: Product of zeroes

LHS: $\alpha\beta\gamma = (3)(-1)(-\frac{1}{3})$

$\alpha\beta\gamma = (-3)(-\frac{1}{3})$

RHS: $-\frac{d}{a}$

Using values from (7) and (4):

From (12) and (13), we see that LHS = RHS. The third relationship is verified.

All the relationships between the zeroes and the coefficients of the polynomial have been verified.

Solution (i): x² – 2x – 8

Given:

The quadratic polynomial $p(x) = x^2 - 2x - 8$.

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To Find:

1. The zeroes of the polynomial.

2. Verify the relationship between the zeroes and the coefficients.

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Solution:

To find the zeroes, we set the polynomial equal to zero:

$x^2 - 2x - 8 = 0$

We factorize the quadratic expression:

$x^2 - 4x + 2x - 8 = 0$

$x(x - 4) + 2(x - 4) = 0$

$(x - 4)(x + 2) = 0$

Setting each factor to zero gives the zeroes:

$x - 4 = 0 \implies x = 4$

$x + 2 = 0 \implies x = -2$

So, the zeroes of the polynomial $x^2 - 2x - 8$ are $4$ and $-2$.

Let $\alpha = 4$ and $\beta = -2$.

Now, we identify the coefficients of the polynomial $x^2 - 2x - 8$. Comparing it with the standard form $ax^2 + bx + c$, we have:

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Verification of Relationship:

1. Sum of zeroes ($\alpha + \beta$):

LHS: $\alpha + \beta = 4 + (-2) = 4 - 2 = 2$

RHS: $-\frac{b}{a} = -\frac{-2}{1} = \frac{2}{1} = 2$

LHS = RHS. The sum of zeroes is verified.

2. Product of zeroes ($\alpha \beta$):

LHS: $\alpha \beta = (4)(-2) = -8$

RHS: $\frac{c}{a} = \frac{-8}{1} = -8$

LHS = RHS. The product of zeroes is verified.

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Solution (ii): 4s² – 4s + 1

Given:

The quadratic polynomial $p(s) = 4s^2 - 4s + 1$.

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To Find:

1. The zeroes of the polynomial.

2. Verify the relationship between the zeroes and the coefficients.

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Solution:

To find the zeroes, we set the polynomial equal to zero:

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

We factorize the quadratic expression. This is a perfect square trinomial:

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

$(2s - 1)^2 = 0$

Taking the square root of both sides:

$2s - 1 = 0$

$2s = 1$

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

This is a repeated zero. So, the zeroes of the polynomial $4s^2 - 4s + 1$ are $\frac{1}{2}$ and $\frac{1}{2}$.

Let $\alpha = \frac{1}{2}$ and $\beta = \frac{1}{2}$.

Now, we identify the coefficients of the polynomial $4s^2 - 4s + 1$. Comparing it with the standard form $as^2 + bs + c$, we have:

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Verification of Relationship:

1. Sum of zeroes ($\alpha + \beta$):

LHS: $\alpha + \beta = \frac{1}{2} + \frac{1}{2} = \frac{1+1}{2} = \frac{2}{2} = 1$

RHS: $-\frac{b}{a} = -\frac{-4}{4} = \frac{4}{4} = 1$

LHS = RHS. The sum of zeroes is verified.

2. Product of zeroes ($\alpha \beta$):

LHS: $\alpha \beta = (\frac{1}{2})(\frac{1}{2}) = \frac{1}{4}$

RHS: $\frac{c}{a} = \frac{1}{4}$

LHS = RHS. The product of zeroes is verified.

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Solution (iii): 6x² – 3 – 7x

Given:

The quadratic polynomial $p(x) = 6x^2 - 3 - 7x$.

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To Find:

1. The zeroes of the polynomial.

2. Verify the relationship between the zeroes and the coefficients.

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Solution:

First, we write the polynomial in the standard form $ax^2 + bx + c$:

$p(x) = 6x^2 - 7x - 3$

To find the zeroes, we set the polynomial equal to zero:

$6x^2 - 7x - 3 = 0$

We factorize the quadratic expression by splitting the middle term ($(-9) \times 2 = -18$ and $-9 + 2 = -7$):

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

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

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

Setting each factor to zero gives the zeroes:

$2x - 3 = 0 \implies 2x = 3 \implies x = \frac{3}{2}$

$3x + 1 = 0 \implies 3x = -1 \implies x = -\frac{1}{3}$

So, the zeroes of the polynomial $6x^2 - 7x - 3$ are $\frac{3}{2}$ and $-\frac{1}{3}$.

Let $\alpha = \frac{3}{2}$ and $\beta = -\frac{1}{3}$.

Now, we identify the coefficients of the polynomial $6x^2 - 7x - 3$. Comparing it with the standard form $ax^2 + bx + c$, we have:

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Verification of Relationship:

1. Sum of zeroes ($\alpha + \beta$):

LHS: $\alpha + \beta = \frac{3}{2} + (-\frac{1}{3}) = \frac{3}{2} - \frac{1}{3} = \frac{3 \times 3 - 1 \times 2}{6} = \frac{9 - 2}{6} = \frac{7}{6}$

RHS: $-\frac{b}{a} = -\frac{-7}{6} = \frac{7}{6}$

LHS = RHS. The sum of zeroes is verified.

2. Product of zeroes ($\alpha \beta$):

LHS: $\alpha \beta = (\frac{3}{2})(-\frac{1}{3}) = -\frac{3}{6} = -\frac{1}{2}$

RHS: $\frac{c}{a} = \frac{-3}{6} = -\frac{1}{2}$

LHS = RHS. The product of zeroes is verified.

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Solution (iv): 4u² + 8u

Given:

The quadratic polynomial $p(u) = 4u^2 + 8u$.

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To Find:

1. The zeroes of the polynomial.

2. Verify the relationship between the zeroes and the coefficients.

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Solution:

To find the zeroes, we set the polynomial equal to zero:

$4u^2 + 8u = 0$

We can factor out the common term $4u$:

$4u(u + 2) = 0$

Setting each factor to zero gives the zeroes:

$4u = 0 \implies u = 0$

$u + 2 = 0 \implies u = -2$

So, the zeroes of the polynomial $4u^2 + 8u$ are $0$ and $-2$.

Let $\alpha = 0$ and $\beta = -2$.

Now, we identify the coefficients of the polynomial $4u^2 + 8u$. Comparing it with the standard form $au^2 + bu + c$, we have:

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Verification of Relationship:

1. Sum of zeroes ($\alpha + \beta$):

LHS: $\alpha + \beta = 0 + (-2) = -2$

RHS: $-\frac{b}{a} = -\frac{8}{4} = -2$

LHS = RHS. The sum of zeroes is verified.

2. Product of zeroes ($\alpha \beta$):

LHS: $\alpha \beta = (0)(-2) = 0$

RHS: $\frac{c}{a} = \frac{0}{4} = 0$

LHS = RHS. The product of zeroes is verified.

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Solution (v): t² – 15

Given:

The quadratic polynomial $p(t) = t^2 - 15$.

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To Find:

1. The zeroes of the polynomial.

2. Verify the relationship between the zeroes and the coefficients.

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Solution:

To find the zeroes, we set the polynomial equal to zero:

$t^2 - 15 = 0$

$t^2 = 15$

Taking the square root of both sides:

$t = \pm \sqrt{15}$

So, the zeroes of the polynomial $t^2 - 15$ are $\sqrt{15}$ and $-\sqrt{15}$.

Let $\alpha = \sqrt{15}$ and $\beta = -\sqrt{15}$.

Now, we identify the coefficients of the polynomial $t^2 - 15$. We can write it as $t^2 + 0t - 15$. Comparing it with the standard form $at^2 + bt + c$, we have:

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Verification of Relationship:

1. Sum of zeroes ($\alpha + \beta$):

LHS: $\alpha + \beta = \sqrt{15} + (-\sqrt{15}) = \sqrt{15} - \sqrt{15} = 0$

RHS: $-\frac{b}{a} = -\frac{0}{1} = 0$

LHS = RHS. The sum of zeroes is verified.

2. Product of zeroes ($\alpha \beta$):

LHS: $\alpha \beta = (\sqrt{15})(-\sqrt{15}) = -(\sqrt{15})^2 = -15$

RHS: $\frac{c}{a} = \frac{-15}{1} = -15$

LHS = RHS. The product of zeroes is verified.

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Solution (vi): 3x² – x – 4

Given:

The quadratic polynomial $p(x) = 3x^2 - x - 4$.

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To Find:

1. The zeroes of the polynomial.

2. Verify the relationship between the zeroes and the coefficients.

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Solution:

To find the zeroes, we set the polynomial equal to zero:

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

We factorize the quadratic expression by splitting the middle term ($(-4) \times 3 = -12$ and $3 - 4 = -1$):

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

$3x(x + 1) - 4(x + 1) = 0$

$(x + 1)(3x - 4) = 0$

Setting each factor to zero gives the zeroes:

$x + 1 = 0 \implies x = -1$

$3x - 4 = 0 \implies 3x = 4 \implies x = \frac{4}{3}$

So, the zeroes of the polynomial $3x^2 - x - 4$ are $-1$ and $\frac{4}{3}$.

Let $\alpha = -1$ and $\beta = \frac{4}{3}$.

Now, we identify the coefficients of the polynomial $3x^2 - x - 4$. Comparing it with the standard form $ax^2 + bx + c$, we have:

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Verification of Relationship:

1. Sum of zeroes ($\alpha + \beta$):

LHS: $\alpha + \beta = -1 + \frac{4}{3} = \frac{-3 + 4}{3} = \frac{1}{3}$

RHS: $-\frac{b}{a} = -\frac{-1}{3} = \frac{1}{3}$

LHS = RHS. The sum of zeroes is verified.

2. Product of zeroes ($\alpha \beta$):

LHS: $\alpha \beta = (-1)(\frac{4}{3}) = -\frac{4}{3}$

RHS: $\frac{c}{a} = \frac{-4}{3}$

LHS = RHS. The product of zeroes is verified.

The general form of a quadratic polynomial whose zeroes are $\alpha$ and $\beta$ is $p(x) = k(x^2 - (\alpha + \beta)x + \alpha\beta)$, where $k$ is any non-zero real number.

Given the sum of zeroes $(\alpha + \beta)$ and the product of zeroes $(\alpha\beta)$, the polynomial can be written as $p(x) = k(x^2 - (\text{Sum of Zeroes})x + (\text{Product of Zeroes}))$.

We can find one such polynomial by taking $k=1$ or a suitable value to clear denominators.

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Solution (i): Sum = $\frac{1}{4}$, Product = $-1$

Given:

Sum of zeroes $= \frac{1}{4}$

Product of zeroes $= -1$

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To Find:

A quadratic polynomial with the given sum and product of zeroes.

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Solution:

Using the formula with $k=1$:

$p(x) = x^2 - (\frac{1}{4})x + (-1)$

$p(x) = x^2 - \frac{1}{4}x - 1$

To get a polynomial with integer coefficients, we can choose $k=4$:

$p(x) = 4(x^2 - \frac{1}{4}x - 1)$

$p(x) = 4x^2 - x - 4$

One quadratic polynomial is $\mathbf{4x^2 - x - 4}$.

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Solution (ii): Sum = $\sqrt{2}$, Product = $\frac{1}{3}$

Given:

Sum of zeroes $= \sqrt{2}$

Product of zeroes $= \frac{1}{3}$

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To Find:

A quadratic polynomial with the given sum and product of zeroes.

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Solution:

Using the formula with $k=1$:

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

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

To get a polynomial with integer coefficients in the denominators, we can choose $k=3$:

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

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

One quadratic polynomial is $\mathbf{3x^2 - 3\sqrt{2}x + 1}$.

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Solution (iii): Sum = $0$, Product = $\sqrt{5}$

Given:

Sum of zeroes $= 0$

Product of zeroes $= \sqrt{5}$

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To Find:

A quadratic polynomial with the given sum and product of zeroes.

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Solution:

Using the formula with $k=1$:

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

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

$p(x) = x^2 + \sqrt{5}$

One quadratic polynomial is $\mathbf{x^2 + \sqrt{5}}$.

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Solution (iv): Sum = $1$, Product = $1$

Given:

Sum of zeroes $= 1$

Product of zeroes $= 1$

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To Find:

A quadratic polynomial with the given sum and product of zeroes.

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Solution:

Using the formula with $k=1$:

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

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

One quadratic polynomial is $\mathbf{x^2 - x + 1}$.

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Solution (v): Sum = $-\frac{1}{4}$, Product = $\frac{1}{4}$

Given:

Sum of zeroes $= -\frac{1}{4}$

Product of zeroes $= \frac{1}{4}$

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To Find:

A quadratic polynomial with the given sum and product of zeroes.

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Solution:

Using the formula with $k=1$:

$p(x) = x^2 - (-\frac{1}{4})x + (\frac{1}{4})$

$p(x) = x^2 + \frac{1}{4}x + \frac{1}{4}$

To get a polynomial with integer coefficients, we can choose $k=4$:

$p(x) = 4(x^2 + \frac{1}{4}x + \frac{1}{4})$

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

One quadratic polynomial is $\mathbf{4x^2 + x + 1}$.

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Solution (vi): Sum = $4$, Product = $1$

Given:

Sum of zeroes $= 4$

Product of zeroes $= 1$

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To Find:

A quadratic polynomial with the given sum and product of zeroes.

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Solution:

Using the formula with $k=1$:

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

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

One quadratic polynomial is $\mathbf{x^2 - 4x + 1}$.