Question

Find the solution of the equation $\sqrt{x^2-16}-(x-4)=\sqrt{x^2-5x+4}$

Answer 1

$\sqrt{x^2-16}-\left(x-4\right)=\sqrt{x^2-5x+4}$

$\sqrt{x^2-16}-\sqrt{x^2-5x+4}=\left(x-4\right)$

$\sqrt{\left(x-4\right)\left(x+4\right)}-\sqrt{\left(x-4\right)\left(x-1\right)}=x-4$

$\sqrt{x-4}\left(\sqrt{x+4}-\sqrt{x-1}\right)=x-4$

Squaring both sides, we get

$\left[x+4+x-1-2\sqrt{x+4}\sqrt{x-1}\right]=\left(x-4\right)$

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

$x+7=2\sqrt{x+4}\sqrt{x-1}$

Squaring both sides, we get

$x^2+49+14x=4\left(x+4\right)\left(x-1\right)$

$x^2+14x+49=4\left(x^2+3x-4\right)$

$x^2+14x+49=4x^2+12x-16$

$-3x^2+2x+65=0$

$3x^2-2x-65=0$

$3x^2+13x-15x-65=0$

$x\left(3x+13\right)-5\left(3x+13\right)=0$

$\left(3x+13\right)\left(x-5\right)=0$

$x=5,-\frac{13}{3}$