Question

Solve $x^2+6x-7=0$ by the method of completing the square

Answer 1

Solution:

Completing the square method:

$x^2+6x=7$

Adding ${[\frac{1}{2}\times \text{coefficient of}x]}^2$ both sides we get,

$x^2+6x+{[\frac{1}{2}\times 6]}^2=7+{[\frac{1}{2}\times 6]}^2$

or, $x^2+6x+3^2=7+3^2$

or, $(x+3{)}^2=16$

or, $x+3=\sqrt{16}$

or, $x+3=\pm 4$

or, $x=4-3$ or $x=-4-3$

or, $x=1$ or $x=-7$