# How do you solve by completing the square for x^2-8x+7?

Jul 7, 2015

You have to take half the $x$-coefficient (=-4) and square it. At the same time you have to balance the 7 that is allready there.

#### Explanation:

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

So we need to add $9$ to both sides:
${x}^{2} - 8 x + 16 = 9 \to$
${\left(x - 4\right)}^{2} = 9 \to$
So: $x - 4 = \sqrt{9} = 3 \to x = 7$
Or: $x - 4 = - \sqrt{9} = - 3 \to x = 1$