Question

Solve the following equation:
$\sqrt{2-x}+\frac{4}{\sqrt{2-x}+3}=2$

Answer 1

$\sqrt{2-x}+\frac{4}{\sqrt{2-x}+3}=2$

$\frac{4}{\sqrt{2-x}+3}+\sqrt{2-x}(\sqrt{2-x}+3)=2$

$4+2-x+3\sqrt{2-x}=2\sqrt{2-x}+66-x-6=-\sqrt{2-x}$

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

$x=\sqrt{2-x}$

$x^2=2-x$

$x^2+x-2=0$

$x^2+2x-x-2=0$

$(x+2)(x-1)=0$

$x=1and-2$

$butx=-2does{n}^{\prime }tsatisfy$

$x=1$