# Question #02cfe

Oct 14, 2016

${x}_{1} = 4 + 4 \sqrt{3}$ and ${x}_{2} = 4 - 4 \sqrt{3}$

#### Explanation:

${x}^{2} - 8 x = 32$

${x}^{2} - 8 x - 32 = 0$

I usually use $\Delta$ to solve any quadratic equation because I'm not good at guessing :)

$\Delta = {b}^{2} - 4 a c$, with $a = 1$, $b = - 8$ and $c = - 32$

$\Delta = {\left(- 8\right)}^{2} - 4 \left(1\right) \left(- 32\right) = 64 + 128 = 192$

$\implies \sqrt{\Delta} = \pm \sqrt{192} = \pm \sqrt{{2}^{6} \times 3} = \pm \sqrt{{2}^{6}} \sqrt{3} = \pm {2}^{3} \sqrt{3}$

$\implies \sqrt{\Delta} = \pm 8 \sqrt{3}$

${x}_{1} = \frac{- b + \sqrt{\Delta}}{2 a}$ and ${x}_{2} = \frac{- b - \sqrt{\Delta}}{2 a}$

${x}_{1} = \frac{8 + 8 \sqrt{3}}{2} = \frac{4 \textcolor{red}{\cancel{8}} \left(1 + 1 \sqrt{3}\right)}{1 \textcolor{red}{\cancel{2}}} = 4 \left(1 + 1 \sqrt{3}\right) = 4 + 4 \sqrt{3}$

and

${x}_{2} = \frac{8 - 8 \sqrt{3}}{2} = \frac{4 \textcolor{red}{\cancel{8}} \left(1 - 1 \sqrt{3}\right)}{1 \textcolor{red}{\cancel{2}}} = 4 \left(1 - 1 \sqrt{3}\right) = 4 - 4 \sqrt{3}$

Your solution is ${x}_{1} = 4 + 4 \sqrt{3}$ and ${x}_{2} = 4 - 4 \sqrt{3}$

Let me know if you have any question :)

I hope this helps :)