Question

Solve the following quadratic equation by completing square method.
$x^2+8x-48=0$.

Answer 1

$x^2+8x-48=0$

$\Rightarrow x^2+8x+16-64=0$

$\Rightarrow x^2+2.x.4+4^2-64=0$

$\Rightarrow (x+4{)}^2-(8{)}^2=0$

$\Rightarrow (x+4+8)(x+4-8)=0$

$\Rightarrow (x+12)(x-4)=0$

$\therefore x=-12,x=4$