Question

Solve $\sqrt{x^2+4x-21}+\sqrt{x^2-x-6}=\sqrt{6x^2-5x-39}$.

Answer 1

$\sqrt{x^2+4x-21}+\sqrt{x^2-x-6}=\sqrt{6x^2-5x-39}$

$\Rightarrow$ $\sqrt{(x+7)(x-3)}+\sqrt{(x-3)(x+2)}=\sqrt{(x-3)(6x+13)}$

$\Rightarrow$ $\sqrt{(x-3)(x+7)}+\sqrt{(x-3)(x+2)}-\sqrt{(x-3)(6x+13)}=0$

$\Rightarrow$ $\sqrt{x-3}(\sqrt{x+7}+\sqrt{x+2}-\sqrt{6x+13})=0$

$\Rightarrow$ $\sqrt{x-3}=0$

$\Rightarrow$ $x-3=0$

$\Rightarrow$ $x=3$

Now,

$\Rightarrow$ $\sqrt{x+7}+\sqrt{x+2}+\sqrt{6x+13}=0$

$\Rightarrow$ $\sqrt{x+7}+\sqrt{x+2}=-\sqrt{6x+13}$

Squaring both sides,

$\Rightarrow$ $x+7+x+2+2\sqrt{x+7}\sqrt{x+2}=6x+13$

$\Rightarrow$ $4x+4=2\sqrt{x+7}\sqrt{x+2}$

Squaring both the sides,

$\Rightarrow$ $4x^2+4+8x=(x+7)(x+2)$

$\Rightarrow$ $4(x^2+1+2x)=x^2+9x+14$

$\Rightarrow$ $3x^2-x-10=0$

$\Rightarrow$ $(x-2)(3x+5)=0$

$\Rightarrow$ $x=2$ and $x=\frac{-5}{3}$

$\Rightarrow$ $x=3,2,\frac{-5}{3}$