Question

Solve this one...
$\sqrt{2} {x}^{2} - 3x - 2 \sqrt{2} = 0$

By Using the method of completing the square.

Answer 1

Solution :

√2x² - 3x - 2√2 = 0

=> Divide each term by √2 ,we get

=> x² - ( 3/√2 )x - 2 = 0

=> x² - 2(x)(3/2√2) = 2

=> x² - 2(x)(3/2√2) + ( 3/2√2)²=2+(3/2√2)²

=> ( x - 3/2√2)² = 2 + 9/8

=> ( x - 3/2√2)² = ( 16 + 9 )/8

=> ( x - 3/2√2)² = 25/8

=> x - 3/2√2 = ± 5/2√2

=> x = 3/2√2 ± 5/2√2

=> x = ( 3 ± 5)/2√2

=> x = 8/2√2 or x = -2/2√2

=> x = 4/√2 or x = -1/√2

=> x = 2√2 or x = -1/√2

•••••

Answer 2

x =-1/√2

hope it helps i