---
title: "NCERT Solutions for Class 10 Maths Chapter 2 Exercise 2.2"
url: https://www.swavid.com/maths/class/10/chapter/polynomials/ncert-solutions/exercise-2-2
dateModified: 2026-10-07T15:48:01+00:00
---

# NCERT Solutions for Class 10 Maths Chapter 2 Exercise 2.2

Chapter 2: Polynomials. Every question from Exercise 2.2, with full working and the final answer.

Free PDF (5 pages): https://www.swavid.com/api/seo/pdf/ncert/maths/class-10/swavid-ncert-solutions-class-10-maths-chapter-2-polynomials-68994c428f.pdf

## EXERCISE 2.2

### Question 1

*3 marks · Short answer*

Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.
(i) $x^2 - 2x - 8$
(ii) $4s^2 - 4s + 1$
(iii) $6x^2 - 3 - 7x$
(iv) $4u^2 + 8u$
(v) $t^2 - 15$
(vi) $3x^2 - x - 4$

**Part (i)**

1. $x^2 - 2x - 8 = (x - 4)(x + 2)$
2. The zeroes are $4$ and $-2$
3. Sum of zeroes = $4 + (-2) = 2 = -\frac{-2}{1}$
4. Product of zeroes = $4 \times (-2) = -8 = \frac{-8}{1}$

Answer (i): Zeroes: $4, -2$

**Part (ii)**

1. $4s^2 - 4s + 1 = (2s - 1)^2$
2. The zeroes are $\frac{1}{2}$ and $\frac{1}{2}$
3. Sum = $\frac{1}{2} + \frac{1}{2} = 1 = -\frac{-4}{4}$
4. Product = $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4} = \frac{1}{4}$

Answer (ii): Zeroes: $\frac{1}{2}, \frac{1}{2}$

**Part (iii)**

1. $6x^2 - 7x - 3 = (3x + 1)(2x - 3)$
2. The zeroes are $-\frac{1}{3}$ and $\frac{3}{2}$
3. Sum = $-\frac{1}{3} + \frac{3}{2} = \frac{7}{6} = -\frac{-7}{6}$
4. Product = $(-\frac{1}{3}) \times \frac{3}{2} = -\frac{1}{2} = \frac{-3}{6}$

Answer (iii): Zeroes: $-\frac{1}{3}, \frac{3}{2}$

**Part (iv)**

1. $4u^2 + 8u = 4u(u + 2)$
2. The zeroes are $0$ and $-2$
3. Sum = $0 + (-2) = -2 = -\frac{8}{4}$
4. Product = $0 \times (-2) = 0 = \frac{0}{4}$

Answer (iv): Zeroes: $0, -2$

**Part (v)**

1. $t^2 - 15 = (t - \sqrt{15})(t + \sqrt{15})$
2. The zeroes are $\sqrt{15}$ and $-\sqrt{15}$
3. Sum = $\sqrt{15} - \sqrt{15} = 0 = -\frac{0}{1}$
4. Product = $\sqrt{15} \times (-\sqrt{15}) = -15 = \frac{-15}{1}$

Answer (v): Zeroes: $\sqrt{15}, -\sqrt{15}$

**Part (vi)**

1. $3x^2 - x - 4 = (3x - 4)(x + 1)$
2. The zeroes are $\frac{4}{3}$ and $-1$
3. Sum = $\frac{4}{3} - 1 = \frac{1}{3} = -\frac{-1}{3}$
4. Product = $\frac{4}{3} \times (-1) = -\frac{4}{3} = \frac{-4}{3}$

Answer (vi): Zeroes: $\frac{4}{3}, -1$

**Answer:** Zeroes and relationship verified for all six quadratic polynomials.

> Common mistake: Incorrect signs while splitting the middle term or writing coefficients.

### Question 2

*3 marks · Short answer*

Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.
(i) $\frac{1}{4}, -1$
(ii) $\sqrt{2}, \frac{1}{3}$
(iii) $0, \sqrt{5}$
(iv) $1, 1$
(v) $-\frac{1}{4}, \frac{1}{4}$
(vi) $4, 1$

**Part (i)**

1. Let the quadratic polynomial be $ax^2 + bx + c$ and its zeroes be $\alpha$ and $\beta$.
2. Given sum of zeroes $\alpha + \beta = \frac{1}{4}$ and product of zeroes $\alpha\beta = -1$.
3. A quadratic polynomial is given by $k[x^2 - (\alpha + \beta)x + \alpha\beta] = k\left[x^2 - \frac{1}{4}x - 1\right]$.
4. For $k = 4$, the polynomial is $4x^2 - x - 4$.

Answer (i): $x^2 - \frac{1}{4}x - 1$ (or $4x^2 - x - 4$)

**Part (ii)**

1. Given sum of zeroes $\alpha + \beta = \sqrt{2}$ and product of zeroes $\alpha\beta = \frac{1}{3}$.
2. A quadratic polynomial is given by $k[x^2 - (\alpha + \beta)x + \alpha\beta] = k\left[x^2 - \sqrt{2}x + \frac{1}{3}\right]$.
3. For $k = 3$, the polynomial is $3x^2 - 3\sqrt{2}x + 1$.

Answer (ii): $x^2 - \sqrt{2}x + \frac{1}{3}$ (or $3x^2 - 3\sqrt{2}x + 1$)

**Part (iii)**

1. Given sum of zeroes $\alpha + \beta = 0$ and product of zeroes $\alpha\beta = \sqrt{5}$.
2. A quadratic polynomial is given by $k[x^2 - (\alpha + \beta)x + \alpha\beta] = k[x^2 - 0x + \sqrt{5}]$.
3. For $k = 1$, the polynomial is $x^2 + \sqrt{5}$.

Answer (iii): $x^2 + \sqrt{5}$

**Part (iv)**

1. Given sum of zeroes $\alpha + \beta = 1$ and product of zeroes $\alpha\beta = 1$.
2. A quadratic polynomial is given by $k[x^2 - (\alpha + \beta)x + \alpha\beta] = k[x^2 - x + 1]$.
3. For $k = 1$, the polynomial is $x^2 - x + 1$.

Answer (iv): $x^2 - x + 1$

**Part (v)**

1. Given sum of zeroes $\alpha + \beta = -\frac{1}{4}$ and product of zeroes $\alpha\beta = \frac{1}{4}$.
2. A quadratic polynomial is given by $k[x^2 - (\alpha + \beta)x + \alpha\beta] = k\left[x^2 - \left(-\frac{1}{4}\right)x + \frac{1}{4}\right]$.
3. For $k = 4$, the polynomial is $4x^2 + x + 1$.

Answer (v): $x^2 + \frac{1}{4}x + \frac{1}{4}$ (or $4x^2 + x + 1$)

**Part (vi)**

1. Given sum of zeroes $\alpha + \beta = 4$ and product of zeroes $\alpha\beta = 1$.
2. A quadratic polynomial is given by $k[x^2 - (\alpha + \beta)x + \alpha\beta] = k[x^2 - 4x + 1]$.
3. For $k = 1$, the polynomial is $x^2 - 4x + 1$.

Answer (vi): $x^2 - 4x + 1$

**Answer:** Quadratic polynomials found for all six cases.

> Common mistake: Swapping the signs of the sum of zeroes while substituting into the general formula $x^2 - (\alpha + \beta)x + \alpha\beta$.

## Related pages

- [All Chapter 2 NCERT solutions](https://www.swavid.com/maths/class/10/chapter/polynomials/ncert-solutions)
- [Exercise 2.1](https://www.swavid.com/maths/class/10/chapter/polynomials/ncert-solutions/exercise-2-1)

Solutions written by SwaVid, a personal AI tutor for Class 6 to 10 Maths and Science. Practise this chapter free: https://www.swavid.com/start/student