Question

Find the roots of the following quadratic equation (if they exist) by the method of completing the square.
x2-(√2+1)x+ √2=0

Answer 1

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