# How do you solve using the completing the square method x^2 – 8x = –11?

Mar 2, 2016

My answer was meant to help rapidly. Stress can make students unable to process explanations. Now on to the explanation.

If you have an equation of the type ${x}^{2} = b$,

you know how to solve. $x = \pm \sqrt{b}$

What if you have terms linear in x too? Well this is your lucky day.
For you know that ${\left(x + a\right)}^{2} = {x}^{2} + 2 a x + {a}^{2}$.
If you stare at this equation long enough, you see that the right hand side has a term in x, but the left hand side is a square. No x. And if it's a square, you know how to solve, right?

So all you need is to work BACKWARD. From the right hand side to the left hand side.

Let's do it.

The linear term in our equation is $- 8 x$. The linear term in the square is $2 a x$, so a = -4.

This gives you
${\left(x - 4\right)}^{2}$ = ${x}^{2} - 8 x + 16$

So that

${x}^{2} - 8 x = {\left(x - 4\right)}^{2} - 16$

${\left(x - 4\right)}^{2} - 16 = - 11$

move 16 to the other side

${\left(x - 4\right)}^{2} = 5$

This is an equation of the second degree. It has two roots, since 5 is positive.

These are $x - 4 = \sqrt{5} , x - 4 = - \sqrt{5}$

Back to x. Two solutions

$x = 4 + \sqrt{5}$
$x = 4 - \sqrt{5}$
And you're done :)