---
title: "NCERT Solutions for Class 10 Maths Chapter 4 Exercise 4.3"
url: https://www.swavid.com/maths/class/10/chapter/quadratic-equations/ncert-solutions/exercise-4-3
dateModified: 2026-10-07T15:49:35+00:00
---

# NCERT Solutions for Class 10 Maths Chapter 4 Exercise 4.3

Chapter 4: Quadratic Equations. Every question from Exercise 4.3, with full working and the final answer.

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## EXERCISE 4.3

### Question 1

*3 marks · Short answer*

Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:
(i) $2x^2 - 3x + 5 = 0$
(ii) $3x^2 - 4\sqrt{3}x + 4 = 0$
(iii) $2x^2 - 6x + 3 = 0$

**Part (i)**

1. Here $a = 2, b = -3, c = 5$.
2. Discriminant $b^2 - 4ac = (-3)^2 - 4(2)(5) = 9 - 40 = -31$.
3. Since $b^2 - 4ac < 0$, the given quadratic equation has no real roots.

Answer (i): No real roots

**Part (ii)**

1. Here $a = 3, b = -4\sqrt{3}, c = 4$.
2. Discriminant $b^2 - 4ac = (-4\sqrt{3})^2 - 4(3)(4) = 48 - 48 = 0$.
3. Since $b^2 - 4ac = 0$, the equation has two equal real roots: $x = \frac{-b}{2a} = \frac{4\sqrt{3}}{2(3)} = \frac{2}{\sqrt{3}}$.

Answer (ii): $x = \frac{2}{\sqrt{3}}, \frac{2}{\sqrt{3}}$

**Part (iii)**

1. Here $a = 2, b = -6, c = 3$.
2. Discriminant $b^2 - 4ac = (-6)^2 - 4(2)(3) = 36 - 24 = 12 > 0$.
3. Since $b^2 - 4ac > 0$, the equation has two distinct real roots given by $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
4. Using the quadratic formula, $x = \frac{6 \pm \sqrt{12}}{4} = \frac{6 \pm 2\sqrt{3}}{4} = \frac{3 \pm \sqrt{3}}{2}$.

Answer (iii): $x = \frac{3 + \sqrt{3}}{2}, \frac{3 - \sqrt{3}}{2}$

**Answer:** The nature of roots and their values are found for each equation.

> Common mistake: Forgetting to simplify the radical in the numerator or making sign errors while calculating the discriminant.

### Question 2

*3 marks · Short answer*

Find the values of $k$ for each of the following quadratic equations, so that they have two equal roots.
(i) $2x^2 + kx + 3 = 0$
(ii) $kx(x - 2) + 6 = 0$

**Part (i)**

1. The given equation is $2x^2 + kx + 3 = 0$. Here $a = 2, b = k, c = 3$.
2. For two equal roots, the discriminant must be zero, so $b^2 - 4ac = 0$.
3. Substituting the values, $k^2 - 4(2)(3) = 0$, which gives $k^2 = 24$.
4. Therefore, $k = \pm\sqrt{24} = \pm 2\sqrt{6}$.

Answer (i): $k = \pm 2\sqrt{6}$

**Part (ii)**

1. The given equation is $kx(x - 2) + 6 = 0$, which can be rewritten as $kx^2 - 2kx + 6 = 0$.
2. Here $a = k, b = -2k, c = 6$.
3. For two equal roots, $b^2 - 4ac = 0$, so $(-2k)^2 - 4(k)(6) = 0$.
4. This gives $4k^2 - 24k = 0$, or $4k(k - 6) = 0$. Since $k \neq 0$ for a quadratic equation, we get $k = 6$.

Answer (ii): $k = 6$

**Answer:** Values of $k$ are found using the condition $b^2 - 4ac = 0$.

> Common mistake: Forgetting that $k$ cannot be zero in the second part since the coefficient of $x^2$ is $k$.

### Question 3

*3 marks · Short answer*

Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is $800\text{ m}^2$? If so, find its length and breadth.

**Solution**

1. Let the breadth of the rectangular mango grove be $x\text{ m}$.
2. Then the length of the grove is $2x\text{ m}$.
3. The area of the rectangular grove is given as $800\text{ m}^2$, so $(2x)(x) = 800$.
4. Simplifying, $2x^2 = 800$, which gives $x^2 = 400$, or $x^2 - 400 = 0$.
5. Solving for $x$, we get $x = \pm 20$. Since breadth cannot be negative, $x = 20$.
6. Thus, breadth $= 20\text{ m}$ and length $= 2(20) = 40\text{ m}$.

**Answer:** Yes, it is possible. Length = $40\text{ m}$, breadth = $20\text{ m}$.

> Common mistake: Not discarding the negative value of $x$ when solving $x^2 = 400$.

### Question 4

*3 marks · Short answer*

Is the following situation possible? If so, determine their present ages. The sum of the ages of two friends is $20$ years. Four years ago, the product of their ages in years was $48$.

**Solution**

1. Let the present age of one friend be $x$ years.
2. Then the present age of the other friend is $(20 - x)$ years.
3. Four years ago, their ages were $(x - 4)$ years and $(20 - x - 4) = (16 - x)$ years respectively.
4. According to the question, the product of their ages 4 years ago was 48, so $(x - 4)(16 - x) = 48$.
5. Expanding and simplifying, $16x - x^2 - 64 + 4x = 48$, which gives $-x^2 + 20x - 112 = 0$, or $x^2 - 20x + 112 = 0$.
6. Checking the discriminant, $b^2 - 4ac = (-20)^2 - 4(1)(112) = 400 - 448 = -48 < 0$.
7. Since the discriminant is negative, there are no real roots, meaning this situation is not possible.

**Answer:** Not possible

> Common mistake: Jumping to conclusions without calculating the discriminant to check the existence of real roots.

### Question 5

*3 marks · Short answer*

Is it possible to design a rectangular park of perimeter $80\text{ m}$ and area $400\text{ m}^2$? If so, find its length and breadth.

**Solution**

1. Let the breadth of the rectangular park be $x\text{ m}$.
2. The perimeter of the park is given as $80\text{ m}$, so $2(\text{length} + \text{breadth}) = 80$, which means $\text{length} + \text{breadth} = 40$.
3. Thus, the length of the park is $(40 - x)\text{ m}$.
4. The area of the park is $400\text{ m}^2$, so $x(40 - x) = 400$.
5. Simplifying, $40x - x^2 = 400$, which gives $x^2 - 40x + 400 = 0$.
6. Solving this quadratic equation using factorisation or formula, $(x - 20)^2 = 0$, which gives $x = 20$.
7. Thus, breadth $= 20\text{ m}$ and length $= 40 - 20 = 20\text{ m}$. The park is a square of side $20\text{ m}$.

**Answer:** Yes, it is possible. Length = $20\text{ m}$, breadth = $20\text{ m}$.

> Common mistake: Assuming a rectangle cannot have equal sides (a square is a special case of a rectangle).

## Related pages

- [All Chapter 4 NCERT solutions](https://www.swavid.com/maths/class/10/chapter/quadratic-equations/ncert-solutions)
- [Exercise 4.1](https://www.swavid.com/maths/class/10/chapter/quadratic-equations/ncert-solutions/exercise-4-1)
- [Exercise 4.2](https://www.swavid.com/maths/class/10/chapter/quadratic-equations/ncert-solutions/exercise-4-2)

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