# NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.2

NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.2 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.2.

• Polynomials Class 10 Ex 2.1
• Polynomials Class 10 Ex 2.2
• Polynomials Class 10 Ex 2.3
• Polynomials Class 10 Ex 2.4

## NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.2

Ex 2.2 Class 10 Maths Question 1.
Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and their coefficients:
(i) x2 – 2x – 8
(ii) 4s2 – 4s + 1
(iii) 6x2 – 3 – 7x
(iv) 4u2 + 8u
(v) t2 – 15
(vi) 3x2 – x – 4
Solution:
(i)
x2 – 2x – 8 = x2 – 4x + 2x – 8
= x(x – 4) + 2(x – 4)
= (x + 2) (x – 4)
Either x + 2 = 0 or x – 4 = 0
⇒  x = -2 or x = 4
Hence, zeroes of this polynomial are -2 and 4.
Verification:
Sum of the zeroes = (-2) + (4) = 2
$\frac { (-2) }{ 1 }$ = $\frac { -b }{ a }$
Product of zeroes = (-2) (4) = -8 =  $\frac { -8 }{ 1 }$
=  $\frac { c }{ a }$
Hence verified.

(ii) 4s2 – 4s + 1 = (2s – l)2 = (2s – l)(2s – 1)
Either 2s – 1 = 0 or 2s – 1 = 0
i.e. , s =  $\frac { 1 }{ 2 }$ , $\frac { 1 }{ 2 }$
Hence, the two Zeroes are $\frac { 1 }{ 2 }$ and $\frac { 1 }{ 2 }$
Verification:

Hence verified.

(iii) 6x2 – 3 – 7x = 6x2 – 7x – 3
= 6x2 – 9x + 2x – 3
= 3x (2x – 3) + 1(2x – 3)
= (2x – 3) (3x + 1)
Either 2x – 3 = 0 or 3x+1 = 0

Hence verified.

(iv) 4u2 + 8u  ⇒  4u(u + 2)
Either 4u = 0 or u + 2 = 0
⇒ u = 0 or u = -2
Hence, the two zeroes are 0 and -2.
Verification:
Sum of the zeroes = 0 + (-2) = -2
$\frac { -8 }{ 4 }$ = $\frac { -b }{ a }$
Product of zeroes = 0 x (-2) = 0 =  $\frac { c }{ a }$
Hence verified.

(v) t2 – 15 = t2 – ($\sqrt{15}$)2
= (t + ($\sqrt{15}$) (t- ($\sqrt{15}$)
Either t + ($\sqrt{15}$ = 0 or t – ($\sqrt{15}$ = 0
⇒ t = -($\sqrt{15}$ or t = ($\sqrt{15}$
Hence, the two zeroes are -($\sqrt{15}$ and + $\sqrt{15}$.
Verification:
Sum of the zeroes = -($\sqrt{15}$ + $\sqrt{15}$ = 0
$\frac { -b }{ a }$
Product of zeroes = –$\sqrt{15}$ x $\sqrt{15}$ = -15
$\frac { c }{ a }$
Hence verified.

(vi) 3x2 – x – 4 = 3x2 – 4x + 3x – 4
=  x(3x – 4) + l(3x 4)
= (x + 1) (3x – 4)
Either x + 1 = 0 or 3x-4 = 0
⇒  x = -1 or x =  $\frac { 4 }{ 3 }$
Verification:
Sum of the zeroes = -1 + $\frac { 4 }{ 3 }$ =  $\frac { 1 }{ 3 }$ =  $\frac { -b }{ a }$
Product of zeroes = -1 x $\frac { 4 }{ 3 }$ =  $\frac { -4 }{ 3 }$ =  $\frac { c }{ a }$
Hence verified.

Ex 2.2 Class 10 Maths Question 2.
Find a quadratic polynomial each with the given numbers as the sum and product of zeroes respectively:

Solution:
(i) Let the zeroes of polynomial be α and β.
Then, α + β = $\frac { 1 }{ 4 }$ and αβ = -1
∴ Required polynomial is given by,
x2 – (α + β)x + αβ = x2 – $\frac { 1 }{ 4 }$x + (-1)
= x2 – $\frac { 1 }{ 4 }$x – 1
= 4x2 – x – 4

(ii) Let the zeroes of polynomial be α and β.
Then, α + β= √2 and αβ = $\frac { 1 }{ 3 }$
∴  Required polynomial is:
x2 – (α + β)x + αβ = x2 – √2x + $\frac { 1 }{ 3 }$
= 3x2 – 3√2x + 1

(iii) Let the zeroes of the polynomial be α and β.
Then, α + β = 0 and αβ = √5
∴ Required polynomial
= x2 – (α + β)x + αβ
= x2– 0 x x + √5 = x2 + √5

(iv) Let the zeroes of the polynomial be α and β.
Then, α + β = 1 and αβ = 1.
∴  Required polynomial
= x2 – (α + β)x + αβ
= x2 – x + 1

(v) Let the zeroes of the polynomial be α and β.
Then, α + β = – $\frac { 1 }{ 4 }$ and αβ = $\frac { 1 }{ 4 }$
∴ Required polynomial
= x2 – (α + β)x + αβ
= x2 – (- $\frac { 1 }{ 4 }$ ) + $\frac { 1 }{ 4 }$
=  4x2 + x + 1 = 0

(vi) Let the zeroes of the polynomial be α and β.
Then, α + β = 4 and αβ = 1.
∴ Required polynomial = x2 -(α + β)x + αβ
= x2 – 4x + 1

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