Question

The number of values of $x$ which satisfy the equation $\sqrt{x+5}+\sqrt{x+21}=\sqrt{6x+40}$ is

Answer 1

Given equation
$\sqrt{x+5}+\sqrt{x+21}=\sqrt{6x+40}$
For the square root to exist,
$x+5\ge 0\Rightarrow x\ge -5x+21\ge 0\Rightarrow x\ge -216x+40\ge 0\Rightarrow x\ge -\frac{40}{6}\therefore x\ge -5$
Squaring both sides,
$x+5+x+21+2\sqrt{(x+5)(x+21)}=6x+40$
$\Rightarrow \sqrt{(x+5)(x+21)}=2x+7$
Squaring both sides, we get
$\Rightarrow (x+5)(x+21)=(2x+7{)}^2\Rightarrow x^2+26x+105=4x^2+28x+49\Rightarrow 3x^2+2x-56=0\Rightarrow (x-4\left)\right(3x+14)=0$
$\Rightarrow x=4,-\frac{14}{3}$
Checking validity for both the roots,
When $x=4$
$\Rightarrow 3+5=8\to \text{satisfy the given equation}\text{When}x=-\frac{14}{3}\Rightarrow \sqrt{\frac{1}{3}}+\sqrt{\frac{49}{3}}=\sqrt{12}\to \text{Given equation is not satisfied}$

Hence, $x=4$ is the only root.