Question

Solve the following equation:
$\sqrt{x+3}+\sqrt{x+4}=\sqrt{x+2}+\sqrt{x+7}$

Answer 1

Squaring on both sides

$⟹2x+7+2\sqrt{(x^2+7x+10)}=2x+9+\sqrt{(x^2+9x+14)}$

Simplifying and squaring again

$⟹x^2+7x+10$ $=$$(x^2+9x+15)+$$2$$\sqrt{(x^2+9x+14)}$

$⟹2x+5=-2\sqrt{(x^2+9x+14)}$

Squaring on both sides

$4x^2+20x+25=4x^2+36x+56$

$⟹x=1.9$ (approx)