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
| Board | CBSE |
| Textbook | NCERT |
| Class | Class 10 |
| Subject | Maths |
| Chapter | Chapter 2 |
| Polynomials | |
| Exercise | Ex 2.2 |
| Number of Questions Solved | 2 |
| Category | NCERT 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:
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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