Question

Solve the following equations:
$\sqrt{4x^2-7x-15}-\sqrt{x^2-3x}=\sqrt{x^2-9}$.

Answer 1

$\sqrt{{4x}^2-7x-15}-\sqrt{x^2-3x}=\sqrt{x^2-9}$

$\sqrt{{4x}^2-7x-15}-\sqrt{x^2-3x}-\sqrt{x^2-9}=0$

$\sqrt{{4x}^2-12x+5x-15}-\sqrt{x(x-3)}-\sqrt{(x-3)(x+3)}=0$

$\sqrt{4x(x-3)+5(x-3)}-\sqrt{x(x-3)}-\sqrt{(x-3)(x+3)}=0$

$\sqrt{(4x+5)(x-3)}-\sqrt{x(x-3)}-\sqrt{(x-3)(x+3)}=0$

$\sqrt{x-3}\left(\sqrt{4x+5}\right)-\sqrt{x}-\sqrt{x+3})=0$

$\sqrt{x-3}=0$

$\Rightarrow x=3$

and $\sqrt{4x+5}-\sqrt{x}-\sqrt{x+3}=0$

$\Rightarrow \sqrt{4x+5}-\sqrt{x}=\sqrt{x+3}$

Squaring both sides

$4x+5+x-2\sqrt{x}\sqrt{4x+5}=x+34x+5+x-x-3=2\sqrt{x}\sqrt{4x+5}4x+2=2\sqrt{x}\sqrt{4x+5}2x+1=\sqrt{x}\sqrt{4x+5}$

Again squaring both sides

${4x}^2+1+4x={4x}^2+5x\Rightarrow x=1$

So, the values of $x$ are $1$ and $3$