# How do you solve x^2 - 6x - 46 = -5 by completing the square?

Feb 15, 2016

See solution below.

#### Explanation:

Put the terms that don't have any x's to one side of the equation and then complete the square like you do to convert quadratic functions from standard form to vertex form.

${x}^{2} - 6 x = 41$

$1 \left({x}^{2} - 6 x + n - n\right) = 41$

$n = {\left(\frac{b}{2}\right)}^{2}$

$n = {\left(- \frac{6}{2}\right)}^{2}$

$n = 9$

$1 \left({x}^{2} - 6 x + 9\right) - 9 = 41$

Factor ${x}^{2} - 6 x + 9$ as a perfect square trinomial.

$1 {\left(x - 3\right)}^{2} = 41 + 9$

${\left(x - 3\right)}^{2} = 50$

The left side of the equation squared equals the right. To get rid of the square on the left, we must take the square of the right.

$\left(x - 3\right) = \pm \sqrt{50}$

$x = 3 \pm \sqrt{25 \times 2}$

$x = 3 \pm 5 \sqrt{2}$

Hopefully this helps!