Our ncert solutions for Exercise 2.2 Class 10 maths 2. Polynomials - ncert solutions - Toppers Study is the best material for English Medium students cbse board and other state boards students.

# Exercise 2.2 Class 10 maths 2. Polynomials - ncert solutions - Toppers Study

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## Exercise 2.2 Class 10 maths 2. Polynomials - ncert solutions - Toppers Study

CBSE board students who preparing for **class 10 ncert solutions maths and Mathematics** solved exercise **chapter 2. Polynomials** available and this helps in upcoming exams
2023-2024.

### You can Find Mathematics solution Class 10 Chapter 2. Polynomials

- All Chapter review quick revision notes for chapter 2. Polynomials Class 10
- NCERT Solutions And Textual questions Answers Class 10 Mathematics
- Extra NCERT Book questions Answers Class 10 Mathematics
- Importatnt key points with additional Assignment and questions bank solved.

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### Exercise 2.2 class 10 Mathematics Chapter 2. Polynomials

Sure! The following topics will be covered in this article

- Exercise 2.2 Class 10 Maths 2. Polynomials - Ncert Solutions - Toppers Study
- Class 10 Ncert Solutions
- Solution Chapter 2. Polynomials Class 10
- Solutions Class 10
- Chapter 2. Polynomials Exercise 2.2 Class 10

## 2. Polynomials

### | Exercise 2.2 |

## Exercise 2.2 Class 10 maths 2. Polynomials - ncert solutions - Toppers Study

**Chapter 2. Polynomials**

### Exercise 2.2 Class 10 maths Chapter 2. Polynomials

**Q1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.**

(i) *x*^{2} – 2*x *– 8

**Solution:**

Using Middle term splitting method:

x^{2} – 4x + 2x – 8 = 0

⇒ x(x – 4) + 2(x – 4) = 0

⇒ x(x – 4) + 2(x – 4) = 0

⇒ (x – 4) (x + 2) = 0

⇒ x – 4 = 0 x + 2 = 0

⇒ x = 4, x = – 2

Zeroes; α = 4 β = – 2

Verifying the relationship between the zeroes and the coefficients.

a = 1, b = – 2, and c = – 8

(ii) 4*s*^{2} – 4*s *+ 1

(iii) 6*x*^{2} – 3 – 7*x*

(iv) 4*u*^{2} + 8*u *

(v) *t*^{2 }– 15

(vi) 3*x*^{2} – *x *– 4

**Solution:**

**(ii) 4 s^{2} – 4s + 1**

**Solution: **

4s^{2}– 4s + 1 = 0

⇒ 4s^{2} – 2s – 2s + 1 = 0

⇒ 2s (2s – 1) –1(2s – 1) = 0

⇒ (2s – 1) (2s – 1) = 0

⇒ 2s – 1= 0, 2s – 1= 0

⇒ 2s = 1, 2s = 1

Coefficients

a = 4, b = – 4, c = 1

Relationship between zeroes and coefficients

L.H.S = R.H.S

L.H.S = R.H.S

Hence verified the relationship between coefficients and the zeroes in both cases.

**(iii) 6 x^{2} – 3 – 7x**

**Solution:**

⇒ 6x^{2 }– 3 – 7x = 0

⇒ 6x^{2 }– 7x – 3 = 0 (After rearranging the equation)

⇒ 6x^{2 }– 9x + 2x – 3 = 0

⇒ 3x (2x – 3) +1 (2x – 3) = 0

⇒ (2x – 3) (3x + 1) = 0

⇒ 2x – 3 = 0, 3x + 1 = 0

⇒ 2x = 3, 3x = - 1

Hence verified the relationship between coefficients and the zeroes in both cases.

**(iv) 4 u^{2} + 8u**

**Solution:** 4u(u + 2) = 0

4u = 0, u + 2 = 0

u = 0, u = – 2

Zeroes: a = 0, b = – 2

Coefficients:

a = 4, b = 8, c = 0

Verifying relationship between zeroes and coefficients

⇒ 0 = 0

⇒ L.H.S = R.H.S

Hence verified, the relationship between coefficients and zeroes in both cases.

**(v) t^{2 }– 15**

**Solution:**

t^{2} – 15 = 0

t^{2} = 15

t = ±√15

t = √15, t = – √15

Zeroes: a = √15, b = – √15

Coefficients a = 1, b = 0, c = – 15

Verifying relationship between zeroes and Coefficients

Hence verified, the relationship between coefficients and zeroes in both cases.

**(vi) 3 x^{2} – x – 4**

**Solution:**

3*x*^{2} – *x *– 4 = 0

⇒ 3x^{2 }+ 3x – 4x – 4 = 0

⇒ 3x (x + 1) – 4 (x + 1) = 0

⇒ (x + 1) (3x – 4) = 0

⇒ x + 1 = 0, 3x – 4 = 0

⇒ x = –1 and 3x = 4

L.H.S= R.H.S

Hence verified, the relationship between coefficients and zeroes in both cases.

**Q2. . Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.**

Hence, required quadratic polynomial is 4x^{2} – x – 4

Hence, required quadratic polynomial is 3x^{2 }–3 x + 1

Hence, required quadratic polynomial is x^{2} + √5

Hence, required quadratic polynomial is x^{2 }– x + 1

Hence, required quadratic polynomial is 4X^{2 }+ x + 1

Hence, required quadratic polynomial is x^{2} – 4 + 1

##### Other Pages of this Chapter: 2. Polynomials

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