# Polynomials (Chapter 2, Exercise 2.2) – Class 10 Maths Solutions NCERT

This page consists of the solutions of the Exercise 2.2, Chapter 2 of Class 10 Mathematics under NCERT Syllabus. The Chapter 2 of Class 10 Mathematics is about the Polynomials. The Exercise 2.2 deals with the questions based on the Relationship between Zeros and Coefficients of a Polynomial. The Exercise 2.2 consists of 2 questions only.

The solution of all the questions of Exercise 2.2 of Class 10 Mathematics is given below.

## Question 1: Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.

### i.$x^2-2x-8$

Solution:

Given, the quadratic polynomial $x^2-2x-8$

Factorising,

$x^2-2x-8$

$=x^2-(4-2)x-8$

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

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

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

The zeros of the quadratic polynomial are,

$x-4=0$

$\Rightarrow x=4$

And $x+2=0$

$\Rightarrow x=-2$

Sum of the zeros$=4+(-2)=4-2=2$

Product of the zeros$=4\times (-2)=-8$

Verification:

Here, $a=1,\, b=-2$ and $c=-8$

Sum of zeros$=\frac{-b}{a}=\frac{-(-2)}{1}=2$

Product of zeros$=\frac{c}{a}=\frac{-8}{1}=-8$

Hence verified.

### ii. $4s^2-4s+1$

Solution:

Given, the quadratic polynomial $4s^2-4s+1$

Factorising,

$4s^2-4s+1$

$=4s^2-(2+2)s+1$

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

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

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

$=(2s-1)^2$

The zeros of the quadratic polynomial are,

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

$2s-1=0$ and $2s-1=0$

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

Sum of the zeros$=\frac{1}{2}+\frac{1}{2}=\frac{2}{2}=1$

Product of the zeros$=\frac{1}{2}\times \frac{1}{2}=\frac{1}{4}$

Verification:

Here, $a=4,\, b=-4$ and $c=1$

Sum of zeros$=\frac{-b}{a}=\frac{-(-4)}{4}=1$

Product of zeros$=\frac{c}{a}=\frac{1}{4}=\frac{1}{4}$

Hence verified.

### iii. $6x^2-3-7x$

Solution:

Given, the quadratic polynomial $6x^2-3-7x=6x^2-7x-3$

Factorising,

$6x^2-7x-3$

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

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

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

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

The zeros of the quadratic polynomial are,

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

$2x=3$ and $3x=-1$

$x=\frac{3}{2}$ and $x=\frac{-1}{3}$

Sum of the zeros$=\frac{3}{2}+\frac{-1}{3}=\frac{7}{6}$

Product of the zeros$=\frac{3}{2}\times \frac{-1}{3}=-\frac{1}{2}$

Verification:

Here, $a=6,\, b=-7$ and $c=-3$

Sum of zeros$=\frac{-b}{a}=\frac{-(-7)}{6}=\frac{7}{6}$

Product of zeros$=\frac{c}{a}=\frac{-3}{6}=-\frac{1}{2}$

Hence verified.

### iv. $4u^2+8u$

Solution:

Given, the quadratic polynomial $4u^2+8u$

Equating the polynomial to zero,

$4u^2+8u=0$

$\Rightarrow 4u(u+2)=0$

The zeros of the quadratic polynomial are,

$4u=0$ and $u+2=0$

$u=0$ and $u=-2$

Sum of the zeros$=0+(-2)=-2$

Product of the zeros$=0\times (-2)=0$

Verification:

Here, $a=4,\, b=8$ and $c=0$

Sum of zeros$=\frac{-b}{a}=\frac{-8}{4}=-2$

Product of zeros$=\frac{c}{a}=\frac{0}{4}=0$

Hence verified.

### v. $t^2-15$

Solution:

Given, the quadratic polynomial $t^2-15$

Equating the polynomial to zero,

$t^2-15=0$

$\Rightarrow t^2=15$

$\Rightarrow t=\pm \sqrt{15}$

$\Rightarrow t=+\sqrt{15},\,-\sqrt{15}$

The zeros of the quadratic polynomial are $+\sqrt{15}$ and $-\sqrt{15}$

Sum of the zeros$=+\sqrt{15}-\sqrt{15}=0$

Product of the zeros$=+\sqrt{15}\times -\sqrt{15}=-15$

Verification:

Here, $a=1,\, b=0$ and $c=-15$

Sum of zeros$=\frac{-b}{a}=\frac{-0}{1}=0$

Product of zeros$=\frac{c}{a}=\frac{-15}{1}=-15$

Hence verified.

### vi. $3x^2-x-4$

Solution:

Given, the quadratic polynomial $3x^2-x-4$

Factorising,

$3x^2-x-4$

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

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

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

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

The zeros of the quadratic polynomial are,

$3x-4=0$ and $x+1=0$

$3x=4$ and $x=-1$

$x=\frac{4}{3}$ and $x=-1$

Sum of the zeros$=\frac{4}{3}+(-1)=\frac{1}{3}$

Product of the zeros$=\frac{4}{3}\times (-1)=-\frac{4}{3}$

Verification:

Here, $a=3,\, b=-1$ and $c=-4$

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

Product of zeros$=\frac{c}{a}=\frac{-4}{3}=-\frac{4}{3}$

Hence verified.

## Question 2: Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.

### i.$\frac{1}{4},\, -1$

Solution:

Given,

$α+β=\frac{1}{4}\,\,\,\,\,……(i)$ and

$α\times β=-1\,\,\,\,\,……(ii)$

We know that,

$α+β=\frac{-b}{a}\,\,\,\,\,……(iii)$ and

$α\times β=\frac{c}{a}\,\,\,\,\,……(iv)$

Comparing equation $(i)$ & $(iii)$ and equation $(ii)$ & $(iv)$

$a=4,\, b=-1$ and $c=-4$

Therefore, the required quadratic polynomial will be,

$ax^2+bx+c$

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

$4x^2-x-4$

### ii. $\sqrt{2},\,\frac{1}{3}$

Solution:

Given,

$α+β=\sqrt{2}\,\,\,\,\,……(i)$ and

$α\times β=\frac{1}{3}\,\,\,\,\,……(ii)$

We know that,

$α+β=\frac{-b}{a}\,\,\,\,\,……(iii)$ and

$α\times β=\frac{c}{a}\,\,\,\,\,……(iv)$

Comparing equation $(i)$ & $(iii)$ and equation $(ii)$ & $(iv)$

$a=3,\, b=-3\sqrt{2}$ and $c=1$

Therefore, the required quadratic polynomial will be,

$ax^2+bx+c$

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

$3x^2-2\sqrt{2}x+1$

### iii. $0,\,\sqrt{5}$

Solution:

Given,

$α+β=0\,\,\,\,\,……(i)$ and

$α\times β=\sqrt{5}\,\,\,\,\,……(ii)$

We know that,

$α+β=\frac{-b}{a}\,\,\,\,\,……(iii)$ and

$α\times β=\frac{c}{a}\,\,\,\,\,……(iv)$

Comparing equation $(i)$ & $(iii)$ and equation $(ii)$ & $(iv)$

$a=1,\, b=0$ and $c=\sqrt{5}$

Therefore, the required quadratic polynomial will be,

$ax^2+bx+c$

$1x^2+(0)x+\sqrt{5}$

$x^2+\sqrt{5}$

### iv. $1,\, 1$

Solution:

Given,

$α+β=1\,\,\,\,\,……(i)$ and

$α\times β=1\,\,\,\,\,……(ii)$

We know that,

$α+β=\frac{-b}{a}\,\,\,\,\,……(iii)$ and

$α\times β=\frac{c}{a}\,\,\,\,\,……(iv)$

Comparing equation $(i)$ & $(iii)$ and equation $(ii)$ & $(iv)$

$a=1,\, b=-1$ and $c=1$

Therefore, the required quadratic polynomial will be,

$ax^2+bx+c$

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

$x^2-x+1$

### v. $-\frac{1}{4},\,\frac{1}{4}$

Solution:

Given,

$α+β=-\frac{1}{4}\,\,\,\,\,……(i)$ and

$α\times β=\frac{1}{4}\,\,\,\,\,……(ii)$

We know that,

$α+β=\frac{-b}{a}\,\,\,\,\,……(iii)$ and

$α\times β=\frac{c}{a}\,\,\,\,\,……(iv)$

Comparing equation $(i)$ & $(iii)$ and equation $(ii)$ & $(iv)$

$a=4,\, b=1$ and $c=1$

Therefore, the required quadratic polynomial will be,

$ax^2+bx+c$

$4x^2+1x+1$

$4x^2-x+1$

### vi. $4,\,1$

Solution:

Given,

$α+β=4\,\,\,\,\,……(i)$ and

$α\times β=1\,\,\,\,\,……(ii)$

We know that,

$α+β=\frac{-b}{a}\,\,\,\,\,……(iii)$ and

$α\times β=\frac{c}{a}\,\,\,\,\,……(iv)$

Comparing equation $(i)$ & $(iii)$ and equation $(ii)$ & $(iv)$

$a=1,\, b=-4$ and $c=1$

Therefore, the required quadratic polynomial will be,

$ax^2+bx+c$

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

$x^2-4x+1$

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