# Class 10 RD Sharma Solutions – Chapter 8 Quadratic Equations – Exercise 8.4

### Question 1: Find the roots of the following quadratic (if they exist) by the method of completing the square: .

**Solution:**

Given:

We have to make the equation a perfect square.

=>

=>

We know that:

=> (

a−b)^{2 }=a^{2}−2×a×b+b^{2}Thus, the equation can be written as:

=>

=>

=>

=>

=>

The RHS is positive, which implies that the roots exist.

=>

=> x = and x=

=>x = and x =

### Question 2: Find the roots of the following quadratic (if they exist) by the method of completing the square: 2x^{2}-7x+3 = 0.

**Solution:**

Given: 2x

^{2}-7x+3 = 0We have to make the equation a perfect square.

=> 2x

^{2}-7x+3 = 0=>

=>

We know that:

=> (a−b)

^{2}=a^{2}−2×a×b+b^{2}Thus, the equation can be written as:

=>

=>

=>

The RHS is positive, which implies that the roots exist.

=>

=> and

=> and

=>x = 3 and

### Question 3: Find the roots of the following quadratic (if they exist) by the method of the completing the square: 3x^{2}+11x+10 = 0.

**Solution:**

Given: 3x

^{2}+11x+10 = 0We have to make the equation a perfect square.

=> 3x

^{2}+11x+10 = 0=>

=>

We know that:

=> (a−b)

^{2}=a^{2}−2×a×b+b^{2}Thus the equation can be written as:

=>

=>

The RHS is positive, which implies that the roots exist.

=>

=> and

=> and

=>and x = -2

### Question 4: Find the roots of the following quadratic (if they exist) by the method of completing the square: 2x^{2}+x-4 =0.

**Solution:**

Given: 2x

^{2}+x-4 =0We have to make the equation a perfect square.

=> 2x

^{2}+x-4 =0=>

=>

We know that:

=> (a−b)

^{2}=a^{2}−2×a×b+b^{2}Thus the equation can be written as:

=>

The RHS is positive, which implies that the roots exist.

=>

=>and

### Question 5: Find the roots of the following quadratic (if they exist) by the method of completing the square: 2x^{2}+x+4 =0.

**Solution:**

Given: 2x

^{2}+x+4 =0We have to make the equation a perfect square.

=> 2x

^{2}+x+4 =0=>

=>

We know that:

=> (a−b)

^{2}=a^{2}−2×a×b+b^{2}Thus the equation can be written as:

=>

=>The RHS is negative, which implies that the roots are not real.

### Question 6: Find the roots of the following quadratic (if they exist) by the method of completing the square: 4x^{2}+4√3+3=0.

**Solution:**

Given: 4x

^{2}+4√3+3=0We have to make the equation a perfect square.

=> 4x

^{2}+4√3+3=0=>

=>

We know that,

=> (a−b)

^{2}=a^{2}−2×a×b+b^{2}Thus the equation can be written as:

=>

=>

=>

The RHS is zero, which implies that the roots exist and are equal.

=>

### Question 7: Find the roots of the following quadratic (if they exist) by the method of the completing the square: .

**Solution:**

Given:

We have to make the equation a perfect square.

=>

=>

=>

We know that,

=> (a−b)

^{2}=a^{2}−2×a×b+b^{2}Thus the equation can be written as:

=>

=>

=>

The RHS is positive, which implies that the roots exist.

=>

=> and

=>and

### Question 8: Find the roots of the following quadratic (if they exist) by the method of completing the square: .

**Solution:**

Given:

We have to make the equation a perfect square.

=>

=>

=>

We know that,

=> (a−b)

^{2}=a^{2}−2×a×b+b^{2}Thus the equation can be written as:

=>

=>

=>

The RHS is positive, which implies that the roots exist.

=>

=> and

=>and

### Question 9: Find the roots of the following quadratic (if they exist) by the method of completing the square: .

**Solution:**

Given:

We have to make the equation a perfect square.

=>

=>

We know that,

=> (a−b)

^{2}=a^{2}−2×a×b+b^{2}Thus the equation can be written as:

=>

=>

=>

The RHS is positive, which implies that the roots exist.

=>

=> and

=>x = √2 and x = 1

### Question 10: Find the roots of the following quadratic equation (if they exist) by the method of completing the square: x^{2}-4ax+4a^{2}-b^{2}=0.

**Solution:**

Given: x

^{2}-4ax+4a^{2}-b^{2}=0We have to make the equation a perfect square.

=> x

^{2}-4ax+4a^{2}-b^{2}=0=> x

^{2}−2×x×2a+(2a)^{2}−b^{2}=0We know that,

=> (a−b)

^{2}=a^{2}−2×a×b+b^{2}Thus the equation can be written as:

=> x

^{2}−2×2a×x+(2a)^{2}=b^{2}=> (x-2a)

^{2}= b^{2}The RHS is positive, which implies that the roots exist.

=> (x-2a) = ±b

=>x= 2a+b and x = 2a-b