How do you solve x^2 - 2x = 32?

Apr 8, 2017

Solutuion : $x = 1 + \sqrt{33} \mathmr{and} x = 1 - \sqrt{33}$

Explanation:

${x}^{2} - 2 x = 32 \mathmr{and} {x}^{2} - 2 x + 1 = 32 + 1 \mathmr{and} {\left(x - 1\right)}^{2} = 33 \mathmr{and} x - 1 = \pm \sqrt{33}$

or $x = 1 + \sqrt{33} \mathmr{and} x = 1 - \sqrt{33}$

Solutuion : $x = 1 + \sqrt{33} \mathmr{and} x = 1 - \sqrt{33}$ [Ans]

Apr 8, 2017

color(blue)(x=1-sqrt33 or color(blue)(x=1+sqrt33

Explanation:

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

$\therefore {x}^{2} - 2 x - 32 = 0$

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

:.x=-(-2)+-sqrt((-2)^2-4(1)(-32))/(2(1)

$\therefore x = \frac{2 \pm \sqrt{4 + 128}}{2}$

$\therefore x = \frac{2 \pm \sqrt{132}}{2}$

$\therefore x = 2 \pm \frac{\sqrt{2 \cdot 2 \cdot 3 \cdot 11}}{2}$

$\therefore x = \frac{2 \pm 2 \sqrt{33}}{2}$

$\therefore x = \frac{2 - 2 \sqrt{33}}{2}$

$\therefore x = \frac{{\cancel{2}}^{\textcolor{b l u e}{1}} \left(1 - \sqrt{33}\right)}{\cancel{2}} ^ \textcolor{b l u e}{1}$

$\therefore x = 1 - \sqrt{33}$

or:

$\therefore x = \frac{2 + 2 \sqrt{33}}{2}$

$\therefore x = {\cancel{2}}^{\textcolor{b l u e}{1}} \frac{1 + \sqrt{33}}{\cancel{2}} ^ \textcolor{b l u e}{1}$

:.color(blue)(x=1+sqrt33