Find the roots of the following quadratic equation (if they exist) by the method of completing the square.
x2-(√2+1)x+ √2=0
Answers
SOLUTION :
Given : x² - (√2 + 1)x + √2 = 0
Shift the constant term on RHS
x² - (√2 + 1)x = - √2
Add square of the ½ of the coefficient of x on both sides
On adding (½ of (√2 + 1))² =(√2 + 1)/2)² both sides
x² - (√2 + 1)x + (√2 + 1)/2)² = - √2 + (√2 + 1)/2)²
Write the LHS in the form of perfect square
(x - (√2 + 1)/2)² = - √2 + [(√2² + 1² + 2√2)/4]
[a² - 2ab + b² = (a - b)² & (a + b)² = a² + 2ab + b² ]
(x - (√2 + 1)/2)² = - √2 + [ (2 + 1 + 2√2)/4]
(x - (√2 + 1)/2)² = - √2 + [ 3 + 2√2)/4]
(x - (√2 + 1)/2)² = [ (- 4√2 + 3 + 2√2)/4]
(x - (√2 + 1)/2)² = ( 3 - 2√2)/4
On taking square root on both sides
(x - (√2 + 1)/2) = √[(3 - 2√2)/4]
(x - (√2 + 1)/2) = √[√2 - 1)²]/2
[√2 - 1)² = √2² + 1² - 2√2 = 2 + 1- 2√2 = 3 - 2√2]
(x - (√2 + 1)/2) = ± [√2 - 1)]/2
On shifting constant term (√2 + 1)/2 to RHS
x = ± [√2 - 1)]/2 + (√2 + 1)/2
x = [(√2 - 1)]/2 + (√2 + 1)/2
[Taking + ve sign]
x = (√2 - 1 + √2 +1)/2
x = 2√2/2 = √2
x = √2
x = - [(√2 - 1)]/2 + (√2 + 1)/2
[Taking - ve sign]
x = (- √2 + 1 + √2 +1)/2
x = 2/2 = 1
x = 1
Hence, the roots of the given equation are √2 & 1