How do you solve using the completing the square method  x^2-8x+7=0?

Sep 20, 2016

$x = 7 \textcolor{w h i t e}{\text{XX")orcolor(white)("XX}} x = 1$

Explanation:

Given
$\textcolor{w h i t e}{\text{XXX")x^2-8x+7=0color(white)("XX}}$Note: I modified the question to include the $\textcolor{b l u e}{= 0}$

Then
$\textcolor{w h i t e}{\text{XXX}} {x}^{2} - 8 x = - 7$

Remembering the general relation:
$\textcolor{w h i t e}{\text{XXX}} {x}^{2} + 2 a x + {a}^{2} = {\left(x + a\right)}^{2}$

If ${x}^{2} - 8 x$ are to be the first two terms of ${x}^{2} + 2 a x + {a}^{2}$
then
$\textcolor{w h i t e}{\text{XXX}} a = - 4$
and
$\textcolor{w h i t e}{\text{XXX}} {a}^{2} = 16$

So we will need to add $16$ (to both sides) to $\underline{\text{complete the square}}$

$\textcolor{w h i t e}{\text{XXX}} {x}^{2} - 8 x + 16 = - 7 + 16$

$\textcolor{w h i t e}{\text{XXX}} {\left(x - 4\right)}^{2} = 9$

$\textcolor{w h i t e}{\text{XXX}} x - 4 = \pm 3$

$\textcolor{w h i t e}{\text{XXX}} x = 4 \pm 3$

$\textcolor{w h i t e}{\text{XXX")x=7color(white)("XX")orcolor(white)("XX}} x = 1$