find the roots of the following quadratic equation if they exist by the method of completing the square 2x²+x+4=1
Answers
SOLUTION :
Given : 2x² + x - 4 = 0
On dividing the whole equation by 2,
(x² + x/2 - 4/2) = 0
(x² + x/2 - 2) = 0
Shift the constant term on RHS
x² + x/2 = 2
Add square of the ½ of the coefficient of x on both sides
On adding (½ of 1/2)² = (1/4)² both sides
x² + x/2 + (1/4)²= 2 + (1/4)²
Write the LHS in the form of perfect square
(x + 1/4)² = 2 + 1/16
[a² + 2ab + b² = (a + b)²]
(x + 1/4)² = (2 × 16 + 1)/16
(x + 1/4)² = (32 + 1)/16
(x + 1/4)² = 33/16
On taking square root on both sides
(x + ¼) = √(33/16)
(x + ¼) = ± √(33)/4
On shifting constant term (1/4) to RHS
x =± √(33)/4 - 1/4
x = √(33)/4 - 1/4
[Taking +ve sign]
x = (√33 - 1)/4
x =± √(33)/4 - 1/4
x = - √(33)/4 - 1/4
[Taking - ve sign]
x = (- √33 - 1)/4
Hence, the roots of the given equation are (√33 - 1)/4 & (- √33 - 1)/4
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