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

May 23, 2015

Given $y = {x}^{2} - 8 x + 7$

Note that if ${x}^{2} - 8 x$ are the first two term of a squared expression ${\left(x + a\right)}^{2}$
then $a = \frac{8}{2} = 4$ and ${a}^{2} = 16$

$y = {x}^{2} - 8 x + 7$

$= {x}^{2} - 8 x + \textcolor{red}{{4}^{2}} + 7 \textcolor{red}{- 16}$

$= {\left(x - 4\right)}^{2} - 9$

Now, the real problem
What do you mean by solve and equation of the form $y =$ and expression in terms of $x$?

I will assume you wish to determine the value(s) of $x$ for which $y = 0$ (although this is not obvious).

If ${\left(x - 4\right)}^{2} - 9 = 0 \left(= y\right)$
then
${\left(x - 4\right)}^{2} = 9$

$x - 4 = \pm \sqrt{9} = \pm 3$

and
$x = 7$ or $x = 1$