# How do you solve (x-3)^2-18=0?

Feb 2, 2016

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

#### Explanation:

The answer that was given above is absolutely correct. However, since the equation is given in the form

${\left(x - a\right)}^{2} + b = 0$,

which is basically an intermediate step of the "completion of the circle" method, I would like to show you an easier way to solve it.

$\textcolor{w h i t e}{x}$

${\left(x - 3\right)}^{2} - 18 = 0$

... add $18$ on both sides...

$\iff {\left(x - 3\right)}^{2} = 18$

Now, you can draw the root, but beware: there are two solutions, the negative and the positive one. (Please be sure that this is clear for you. As an example, for ${x}^{2} = 49$, both $x = 7$ and $x = - 7$ are solutions.)

$\implies x - 3 = \sqrt{18} \text{ or } x - 3 = - \sqrt{18}$

... add $3$ on both sides...

$\implies x = 3 + \sqrt{18} \text{ or } x = 3 - \sqrt{18}$

$x = 3 \pm \sqrt{18} = 3 \pm \sqrt{9 \cdot 2} = 3 \pm \sqrt{9} \cdot \sqrt{2} = 3 \pm 3 \sqrt{2}$

Feb 4, 2016

${\left(x - 3\right)}^{2} - 18 = 0$

Use the formula ${\left(a + b\right)}^{2} = {a}^{2} + 2 a b + {b}^{2}$

$\rightarrow {\left(x - 3\right)}^{2} = {x}^{2} + 2 \left(x\right) \left(- 3\right) + {\left(- 3\right)}^{2} = {x}^{2} - 6 x + 9$

So,

$\rightarrow {x}^{2} - 6 x + 9 - 18 = 0$

$\rightarrow {x}^{2} - 6 x - 9 = 0$

Now this is a Quadratic equation (in form $a {x}^{2} + b {x}^{2} + c$)

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

In this case $a = 1 , b = - 6 , c = - 9$

$\rightarrow x = \frac{- \left(- 6\right) \pm \sqrt{{\left(- 6\right)}^{2} - 4 \left(1\right) \left(- 9\right)}}{2 \left(1\right)}$

$\rightarrow x = \frac{6 \pm \sqrt{36 - \left(- 36\right)}}{2}$

$\rightarrow x = \frac{6 \pm \sqrt{36 + 36}}{2}$

$\rightarrow x = \frac{6 \pm \sqrt{72}}{2}$

$\rightarrow x = \frac{6 \pm \sqrt{36 \cdot 2}}{2}$

$\rightarrow x = \frac{6 \pm 6 \sqrt{2}}{2} = \frac{6}{2} \pm \frac{6 \sqrt{2}}{2}$

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