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.

DescriptionNCERT Maths Solutions
SubjectMathematics
Class10 (X)
Chapter2
Exercise2.2
TopicThe Relationship between Zeros and
the Coefficients of a Polynomial
SyllabusNCERT

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

I hope the solutions of Class 10, Exercise 2.2, Chapter 2 of Mathematics have helped you. If you have any issues/queries related to the NCERT Solutions of Class 10 Mathematics, Exercise 2.2, feel free to contact me at [email protected] or [email protected] or fill the form here.

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