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
BoardCBSE
TextbookNCERT
ClassClass 10
SubjectMaths
ChapterChapter 2
Polynomials
ExerciseEx 2.2
Number of Questions Solved2
CategoryNCERT Solutions

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:
NCERT Solutions for Class 10 Maths Chapter 2 Polynomials e1 2
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
byjus class 10 maths Chapter 2 Polynomials e1 2a
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:
NCERT Solutions for Class 10 Maths Chapter 2 Polynomials e2 2
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

We hope the NCERT Solutions for Class 10 Mathematics Chapter 2 Polynomials Ex 2.2 help you. If you have any query regarding NCERT Solutions for Class 10 Mathematics Chapter 2 Polynomials Ex 2.2 drop a comment below and we will get back to you at the earliest.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

%d bloggers like this:
search previous next tag category expand menu location phone mail time cart zoom edit close