# Find the roots of the following quadratic equations, if they exist, by the method of completing the square:

(i) 2x^{2} - 7x + 3 = 0

(ii) 2x^{2} + x - 4 = 0

(iii) 4x^{2} + 4√3x + 3 = 0

(iv) 2x^{2} + x + 4 = 0

**Solution:**

Steps required to solve a quadratic equation by applying the ‘completing the square’ method are given below:

Let the given quadratic equation be ax^{2} - bx + c = 0

(a) Divide all the terms by the coefficient of x^{2} i.e, a

ax^{2}/a + bx/a + c/a = 0

x^{2} + bx/a + c/a = 0

(b) Move the constant term c/a to the right side of the equation:

x^{2} + bx/a = - c/a

(c) Complete the square on the left side of the equation by adding the square of half of the coefficient of x i.e, b^{2}/4a. Balance the equation by adding the same value to the right side of the equation.

x^{2} + bx/a + b^{2}/4a = - c/a + b^{2}/4a

(i) 2x^{2} - 7x + 3 = 0

x^{2} - 7x/2 + 3/2 = 0 (dividing by 2 on both the sides)

x^{2} - 7x/2 = - 3/2

Since 7/2 ÷ 2 = (7/4), Therefore, (7/4)^{2 } should be added to both sides of the quadratic equation:

x^{2} - 7x/2 + (7/4)^{2} = - 3/2 + (7/4)^{2}

(x - 7/4)^{2} = - 3/2 + (49/16) [Since, a^{2} - 2ab + b^{2} = (a - b)^{2}]

(x - 7/4)^{2 }= (-24 + 49) / 16

(x - 7/4)^{2} = 25/16

(x - 7/4)^{2} = (5/4)^{2}

x - 7/4 = 5/4 and x - 7/4 = - 5/4

x = 5/4 + 7/4 and x = - 5/4 + 7/4

x = 12/4 and x = 2/4

x = 3 and x = 1/2

Roots are: 3, 1/2

(ii) 2x^{2} + x - 4 = 0

x^{2} + x/2 - 2 = 0 (dividing by 2 on both the sides)

x^{2} + x/2 = 2

Since 1/2 ÷ 2 = 1/2 × 1/2 = 1/4, [(1/4)^{2} should be added to both sides]

x^{2} + x/2 + (1/4)^{2} = 2 + (1/4)^{2}

[x + (1/4)]^{2} = 2 + 1/16 [Since, a^{2} + 2ab + b^{2} = (a + b)^{2}]

[x + (1/4)]^{2} = (32 + 1) / 16

[x + (1/4)]^{2} = 33/16

x + 1/4 = √33/4 and x + 1/4 = - √33/4

x = √33/4 - 1/4 and x = - √33/4 - 1/4

x = (√33 - 1)/4 and x = (- √33 - 1)/4

Thus, the roots are (√33 - 1)/4 and (- √33 - 1)/4

(iii) 4x^{2} + 4√3x + 3 = 0

x^{2} + √3x + 3/4 = 0 (dividing by 4 on both the sides)

x^{2} + √3x = - 3/4

x^{2} + √3x + (√3/2)² = - 3/4 + (√3/2)^{2} [(√3/2)^{2} is added on both sides]

[x + (√3/2)]^{2} = - 3/4 + 3/4 [Since, a^{2} + 2ab + b^{2} = (a + b)^{2}]

[x + (√3/2)]^{2} = 0

x = - √3/2 and x = - √3/2

Roots are - √3/2 and - √3/2

(iv) 2x^{2} + x + 4 = 0

x^{2} + x/2 = - 2 (dividing both the sides by 2)

Adding (1/4)^{2 }on both sides of the equation,

x^{2} + x/2 + (1/4)^{2} = - 2 + (1/4)^{2}

[x + (1/4)]^{2} = - 2 + 1/16 [Since, a^{2} + 2ab + b^{2} = (a + b)^{2}]

[x + (1/4)]^{2} = (- 32 + 1)/16

(x + (1/4))^{2} = - 31/16 < 0

Square of any real number can’t be negative.

Hence, real roots don’t exist.

**☛ Check: **Class 10 Maths NCERT Solutions Chapter 4

**Video Solution:**

## Find the roots of the following quadratic equations, if they exist, by the method of completing the square: (i) 2x² - 7 x + 3 = 0 (ii) 2x² + x - 4 = 0 (iii) 4x² + 4√3x + 3 = 0 (iv) 2x² + x + 4 = 0

Class 10 Maths NCERT Solutions Chapter 4 Exercise 4.3 Question 1

**Summary:**

The roots of the quadratic equation by the method of completing the squares are (i) 2x^{2} - 7 x + 3 = 0; x = 3 and x = 1/2, (ii) 2x^{2} + x - 4 = 0; x = (√33 - 1)/4 and x = (- √33 - 1)/4, (iii) 4x^{2} + 4√3x + 3 = 0 ; x = - √3/2 and x = - √3/2, (iv) 2x^{2} + x + 4 = 0; no real roots exist

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