# How do you solve x^2 - 8x + 3 = 0 by completing the square?

Oct 8, 2015

The solutions are $4 \setminus \pm \sqrt{13}$.

#### Explanation:

We must manipulate the expression into something of the form

${x}^{2} + 2 a x + {a}^{2}$

Since the term $2 a x$ equals $- 8 x$, it means that $a = - 4$, and thus ${a}^{2} = 16$. We need $13$ more units for the constant to reach that value, so we can add and subtract:

${x}^{2} - 8 x + 3 + 13 - 13 = 0 \to {x}^{2} - 8 x + 16 = 13$

But now we have that ${x}^{2} - 8 x + 16 = {\left(x - 4\right)}^{2}$, so the equation becomes

${\left(x - 4\right)}^{2} = 13$

Taking square roots of both sides:

$x - 4 = \setminus \pm \setminus \sqrt{13}$

$x = 4 \setminus \pm \setminus \sqrt{13}$