Find the roots of the following quadratic equations, if they exist, by the method of completing the square:
(i) 2x2 - 7x + 3 = 0
(ii) 2x2 + x - 4 = 0
(iii) 4x2 + 4√3x + 3 = 0
(iv) 2x2 + 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 ax2 - bx + c = 0
(a) Divide all the terms by the coefficient of x2 i.e, a
ax2/a + bx/a + c/a = 0
x2 + bx/a + c/a = 0
(b) Move the constant term c/a to the right side of the equation:
x2 + 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, b2/4a. Balance the equation by adding the same value to the right side of the equation.
x2 + bx/a + b2/4a = - c/a + b2/4a
(i) 2x2 - 7x + 3 = 0
x2 - 7x/2 + 3/2 = 0 (dividing by 2 on both the sides)
x2 - 7x/2 = - 3/2
Since 7/2 ÷ 2 = (7/4), Therefore, (7/4)2 should be added to both sides of the quadratic equation:
x2 - 7x/2 + (7/4)2 = - 3/2 + (7/4)2
(x - 7/4)2 = - 3/2 + (49/16) [Since, a2 - 2ab + b2 = (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) 2x2 + x - 4 = 0
x2 + x/2 - 2 = 0 (dividing by 2 on both the sides)
x2 + x/2 = 2
Since 1/2 ÷ 2 = 1/2 × 1/2 = 1/4, [(1/4)2 should be added to both sides]
x2 + x/2 + (1/4)2 = 2 + (1/4)2
[x + (1/4)]2 = 2 + 1/16 [Since, a2 + 2ab + b2 = (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) 4x2 + 4√3x + 3 = 0
x2 + √3x + 3/4 = 0 (dividing by 4 on both the sides)
x2 + √3x = - 3/4
x2 + √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, a2 + 2ab + b2 = (a + b)2]
[x + (√3/2)]2 = 0
x = - √3/2 and x = - √3/2
Roots are - √3/2 and - √3/2
(iv) 2x2 + x + 4 = 0
x2 + x/2 = - 2 (dividing both the sides by 2)
Adding (1/4)2 on both sides of the equation,
x2 + x/2 + (1/4)2 = - 2 + (1/4)2
[x + (1/4)]2 = - 2 + 1/16 [Since, a2 + 2ab + b2 = (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) 2x2 - 7 x + 3 = 0; x = 3 and x = 1/2, (ii) 2x2 + x - 4 = 0; x = (√33 - 1)/4 and x = (- √33 - 1)/4, (iii) 4x2 + 4√3x + 3 = 0 ; x = - √3/2 and x = - √3/2, (iv) 2x2 + x + 4 = 0; no real roots exist
☛ Related Questions:
- Find the roots of the quadratic equations given in Q.1 above by applying the quadratic formula.
- Find the roots of the following equations:(i) x - 1/x = 3, x ≠ 0(ii) 1/(x + 4) - 1/(x - 7) = 11/30, x ≠ - 4, 7
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