Question

The only integer solution of the equation $3\sqrt{x+3}-\sqrt{x-2}=7$ is

Answer 1

We have $3\sqrt{x+3}-\sqrt{x-2}=7$
$\Rightarrow 3\sqrt{x+3}=7+\sqrt{x-2}$
Squaring both of the sides of the equation, we obtain$9x+27=49+x-2+14\sqrt{x-2}$
$\Rightarrow 8x-20=14\sqrt{(x-2)}$
$(4x-10)=7\sqrt{x-2}$
again squaring both sides, we obtain${16x}^2+100-80x=49x-98$
$\Rightarrow {16x}^2-129x+198=0$$\Rightarrow (x-6)(x-\frac{33}{16})=0$
$x_1=6$, and $x_2=\frac{33}{16}$
Hence
$x_1=6$ satisfies the original equation but $x_2=\frac{33}{16}$ is not satisfies the original equation.$\therefore x_2=\frac{33}{16}$ is the extraneous roots.