Question

Solve the following equation:
$\frac{8}{\sqrt{10-2x}}-\sqrt{10-2x}=2$

Answer 1

We have, $\frac{8}{\sqrt{10-2x}}-\sqrt{10-2x}=2$

Let $\sqrt{10-2x}=t$

On putting the above value in given equation, we get

$\frac{8}{t}-t=2$

$\Rightarrow 8-t^2=2t$

$\Rightarrow t^2+2t-8=0$

$\Rightarrow t^2+4t-2t-8=0$

$\Rightarrow (t+4)-2(t+4)=0$

$\Rightarrow (t+4)(t-2)=0$

$\therefore t=2,-4$

When $t=-4$

$\sqrt{10-2x}=-4$

$\Rightarrow 10-2x=16$ [On squaring both sides]

$\Rightarrow -2x=6$

$\therefore x=-3$

When $t=2$

$\sqrt{10-2x}=2$

$\Rightarrow 10-2x=4$ [On squaring both sides]

$\Rightarrow -2x=-6$

$\therefore x=3$

Hence, $x=-3,3$