Question

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

Answer 1

$\text{L}et$ $\sqrt{\frac{3-x}{2+x}}$ $\text{b}e$ $tt+\frac{3}{t}=4t^2-4t+3=0(t-3)(t-1)=0t=1,3\sqrt{\frac{3-x}{2+x}}=1\frac{3-x}{2+x}=13-x=2+xx=\frac{1}{2}\sqrt{\frac{3-x}{2+x}}=3\frac{3-x}{2+x}=93-x=18+9xx=-1.5$