Question

Solve the following equation:
$2x^2+\sqrt{2x^2-4x+12}=4x+8$

Answer 1

$2x^2+\sqrt{2x^2-4x+12}=4x+8$

$\Rightarrow 2x^2-4x+\sqrt{2x^2-4x+12}=8$

Let $2x^2-4x=t$

$\Rightarrow t+\sqrt{t+12}=8$

$\Rightarrow \sqrt{t+12}=8-t$

$t+12={(8-t)}^2$

$t+12=64+t^2-16t$

$t^2-17t+52=0$

$t=\frac{17\pm \sqrt{{17}^2-4\times 52}}{2}=\frac{17\pm \sqrt{289-208}}{2}$

$t=\frac{17\pm \sqrt{81}}{2}=\frac{17\pm 9}{2}$

$t=\frac{17+9}{2}$ or $\frac{17-9}{2}$

$t=\frac{17+9}{2}=\frac{26}{2}=13$ or $t=\frac{17-9}{2}=\frac{8}{2}=4$

$\Rightarrow 2x^2-4x=13$ $2x^2-4x=4$

$\Rightarrow 2x^2-4x-13=0$ $x^2-2x-2=0$

$\Rightarrow x=\frac{4\pm \sqrt{16-4\times 2\times -13}}{4}$ $\frac{2\pm \sqrt{4-4\times 1\times -2}}{2}$

$=\frac{4\pm \sqrt{16+104}}{4}$ $\frac{2\pm \sqrt{12}}{2}$

$=\frac{4\pm \sqrt{120}}{4}$ $\frac{2\pm 2\sqrt{3}}{2}=1\pm \sqrt{3}$

$\Rightarrow x=\frac{4\pm 2\sqrt{30}}{4}$ $x=1\pm \sqrt{3}$

Hence, solved.