Question

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

Answer 1

$\sqrt{x+2}+\sqrt{3-x}=3⟹x-2\ge 0$ and $3-x\ge 0⟹x\in [2,3]$

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

Squaring on both sides

$x+2=9+3-x+6\sqrt{3-x}$

$x-5=3\sqrt{3-x}$

$x^2-10x+25=27-9x$

$x^2-x-2=0⟹x=2,-1$