How do you solve w ^ { 2} - 6w + 9= 13?

Nov 3, 2016

$w = 3 \pm \sqrt{13}$

Explanation:

Given:

${w}^{2} - 6 w + 9 = 13$

Note that the left hand side is already a perfect square trinomial, so we have:

${\left(w - 3\right)}^{2} = 13$

Taking the square root of both sides, allowing for both possible square roots, we have:

$w - 3 = \pm \sqrt{13}$

Then add $3$ to both sides to get:

$w = 3 \pm \sqrt{13}$

Nov 3, 2016

$w = 3 \pm \sqrt{13}$

Explanation:

${w}^{2} - 6 w + 9 = 13$
Rearrange
${w}^{2} - 6 w - 4 = 0$
Either use the formula $w = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
Where $a {w}^{2} + b w + c = 0$
$w = \frac{6 \pm \sqrt{36 - 4 \cdot 1 \cdot - 4}}{2}$
$w = \frac{6 + \sqrt{36 + 16}}{2}$ or $w = \frac{6 - \sqrt{36 + 16}}{2}$
$w = \frac{6 + \sqrt{52}}{2} = 3 + \sqrt{13}$
Or
$w = 3 - \sqrt{13}$

The other way is completing the square ( which is how the formula is derived)
${w}^{2} - 6 w - 4 = 0$
${\left(w - 3\right)}^{2} = {w}^{2} - 6 w + 9$
So ${w}^{2} - 6 w - 4$ can be written ${\left(w - 3\right)}^{2} - 13 = 0$
${\left(w - 3\right)}^{2} = 13$
Take the square root of both side
$w - 3 = \pm 13$
$w = 3 \pm 13$