NCERT Solutions Class 10 Chapter 4 Quadratic Equations Exercise 4.3 (original) (raw)

Last Updated : 23 Jul, 2025

Chapter 4 of the Class 10 NCERT Mathematics textbook, titled "Quadratic Equations" focuses on solving quadratic equations using various methods, including factorization, completing the square, and the quadratic formula. Exercise 4.3 specifically deals with solving quadratic equations by the method of completing the square, helping students understand how to manipulate equations to find their roots effectively.

NCERT Solutions for Class 10 - Chapter 4 Quadratic Equations - Exercise 4.3

This section provides detailed solutions for Exercise 4.3 from Chapter 4 of the Class 10 NCERT Mathematics textbook. The exercise emphasizes solving quadratic equations by completing the square. It offers step-by-step explanations to ensure students grasp the technique and can apply it confidently to solve various quadratic equations.

**Question 1. 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√3x+4=0

****(iii) 2x** 2 -6x+3=0

**Solution:

(i) Given: 2x2-3x+5=0

Here a=2,b=-3 and c=5

\therefore Discriminant, D=b2-4ac

= (-3)2- 4 × 2 × 5)

= 9-40 = -31 < 0

Hence, the roots are imaginary.

(ii) Given: 3x2-4√3x + 4 = 0

Here a=3,b=√3 and c=4

\therefore Discriminant, D=b2-4ac

= (-4√3)2 - (4 × 3 × 4)

= 48 - 48 = 0

Hence, the roots are real and equal.

Using the formula,

x=\frac {-b\pm\sqrt{b^2-4ac}} {2a} , we get

x=\frac {-(-4\sqrt3)\pm\sqrt{(-4\sqrt3)^2-4\times3\times4}} {2\times3}

= \frac {4\sqrt3\pm\sqrt{48-48}} {6}=\frac {4\sqrt3} {6}=\frac {2} {\sqrt3}

Hence, the equal roots are \frac 2 {\sqrt3} and \frac 2 {\sqrt3} .

(iii) Given: 2x2-6x+3=0

Here, a=2,b=-6 and c=3

\therefore Discriminant, D=b2-4ac

= (-6)2 - (4 × 2 × 3)

= 36 - 24 = 12 > 0

Hence, the roots are distinct and real.

Using the formula,

x=\frac {-b\pm\sqrt{b^2-4ac}} {2a} ,we get

x=\frac {-(-6)\pm\sqrt{(-6)^2-4\times2\times3}} {2\times2}

x=\frac {6\pm\sqrt{36-24}} {4}

x=\frac {6\pm\sqrt{12}} {4}

x=\frac {6\pm2\sqrt{3}} {4}

x=\frac {3\pm\sqrt{3}} {2}

Hence, the equal roots are \frac {3+\sqrt{3}} {2}and \frac {3-\sqrt{3}} {2}

**Question 2. Find the values of k for each of the following quadratic equations, so that they have two equal roots.

****(i)** **2x 2 +kx+3

****(ii) kx(x-2)+6=0**

**Solution:

****(i) 2x+kx+3=0**

This equation is of the form ax2+bx+x, where a=2, b=k and c=3.

Discriminant, D=b2-4ac

=k2 - 4 × 2 × 3

=k2 -24

For equal roots D=0

\implies k2-24=0

\implies k2=24

\implies __k_2 = ±24 = ±2√6​

****(ii) kx(x-2)+6=0**

\implies kx2-2kx+6=0

This equation is of the form ax2+bx+c=0, where a=k, b=-2k and c=6.

Discriminant, D=b2-4ac

=(-2k)2 - 4 × k × 6

=4k2-24k

For equal roots D=0

\implies 4k2-24k=0

\implies 4k(k-24)=0

\implies k=0 (not possible) or 4k-24=0

\implies k= 24/4=6

**Question 3. Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m 2 ? If so, find its length and breadth.

**Solution:

Let the breadth of the rectangular mango grove be x m.

Then, the length of the rectangular mango grove will be 2x m.

The Area of the rectangular mango grove=length × breadth

According to the question, we have

x × 2x= 800

\implies 2x2=800

\implies x2=400

\implies x=20

Hence, the rectangular mango grove is possible to design whose length=40 m and breadth=20 m.

**Question 4. 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:

Let the present age of one friend be x years.

Then, the present age of other friend be (20-x) years.

4 years ago, one friend's age was (x-4) years

4 years ago, other friend's age was (20-x-4)=(16-x) years.

According to the question,

(x-4)(16-x)=48

\implies 16x-64-x2+4x=48

\implies x2-20x+112=0

This equation is of the form ax2+bx+c=0,where a=1, b=-20 and c=112.

Discriminant, D=b2-4ac

= (-20)2-4 × 1 × 112 = -48 < 0

Since, there are no real roots.

So the given situation is not possible.

**Question 5. Is it possible to design a rectangular park of perimeter 80 m and area 400 m 2 ? If so, find its length and breadth.

**Solution:

Let the length of the rectangular park be x.

The perimeter of the rectangular park= 2(length + breadth)

\implies 2(x + breadth)=80

\implies breadth=40-x

The area of rectangular park= length × breadth

\implies x(40-x)=400

\implies 40x-x2=400

\implies x2-40x+400=0

\implies x2 -20x-20x+400=0

\implies (x-20)(x-20)=0

\implies x=20

Hence, the rectangular park is possible to design. So, the length of the park is 20m and the breadth = 40-20=20m.

Summary

Chapter 4 of the Class 10 NCERT Mathematics textbook, "Quadratic Equations" teaches students how to solve quadratic equations using various methods. Exercise 4.3 focuses on the method of completing the square, which involves transforming a quadratic equation into a form that can be easily solved by taking the square root of both sides. This method is useful when equations don't factor easily and helps students understand the roots of quadratic equations in a systematic way.

NCERT Solutions Class 10 - Chapter 4 Quadratic Equations - Exercise 4.2

NCERT Solutions Class 10 - Chapter 4 Quadratic Equations - Exercise 4.4