Question

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

• A. $\{4,5,-\frac{13}{3}\}$
• B. $\{4,5\}$
• C. $\left\{4\right\}$
• D. $\{5,-\frac{13}{3}\}$

Answer 1

The correct option is A $\{4,5,-\frac{13}{3}\}$

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

$\Rightarrow \sqrt{(x-4)}\{\sqrt{x+4}-\sqrt{x-4}\}=\sqrt{(x-4)}\times \sqrt{(x-1)}$

$\Rightarrow \sqrt{x-4}=0\Rightarrow x=4$

or

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

Squaring both sides,

$x+4+(x-4)-2\sqrt{x^2-16}=x-1$

$\Rightarrow x+1=2\sqrt{x^2-16}$

Squaring both sides we get

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

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

$\therefore x=\{5,4,-\frac{13}{3}\}$

option $A$ is correct.