Solve:x^2-16-x^2-8 x+16=x^2-5 x+4

Question

Solve$:\sqrt{x^2-16}-\sqrt{x^2-8 x+16}=\sqrt{x^2-5 x+4}$

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Answer

$\sqrt{x^2-16}-\sqrt{x^2-8 x+16}=\sqrt{x^2-5 x+4}$
$\therefore \quad \sqrt{(x+4)(x-4)}-\sqrt{(x-4)^2}=\sqrt{(x-4)(x-1)}$
$\therefore \quad \sqrt{(x+4)(x-4)}-\sqrt{(x-4)^2}-\sqrt{(x-4)(x-1)}=0$
$\therefore \quad \sqrt{x-4}[\sqrt{x+4}-\sqrt{(x-4)}-\sqrt{x-1}]=0$
$\therefore \quad \sqrt{x-4}=0$ or $\sqrt{x+4}-\sqrt{x-4}-\sqrt{x-1}=0$
$\therefore \quad x-4=0$ or $\sqrt{x+4}-\sqrt{x-4}=\sqrt{x-1}$
$\therefore \quad x=4$ or $\sqrt{x+4}-\sqrt{x-4}=\sqrt{x-1}$
Now, $\sqrt{x+4}-\sqrt{x-4}=\sqrt{x-1}$...(i)
$\therefore \quad(\sqrt{x+4}-\sqrt{x-4})^2=x-1 \ldots($ Squaring both sides $)$
$\therefore \quad x+4-2 \sqrt{(x+4)(x-4)}+x-4=x-1$
$\therefore \quad 2 x-2 \sqrt{x^2-16}=x-1$
$\therefore \quad 2 x-x+1=2 \sqrt{x^2-16}$
$\therefore \quad x+1=2 \sqrt{x^2-16}$
$\therefore \quad(x+1)^2=4\left(x^2-16\right) \ldots$ (Squaring both sides)
$\therefore \quad x^2+2 x+1=4 x^2-64$
$\therefore \quad 4 x^2-x^2-2 x-64-1=0$
$\therefore \quad 3 x^2-2 x-65=0$
$\therefore \quad 3 x^2-15 x+13 x-65=0$
$\therefore \quad 3 x(x-5)+13(x-5)=0$
$\therefore \quad(3 x+13)(x-5)=0$
$\therefore \quad 3 x+13=0$ or $x-5=0$
$\therefore \quad x=\frac{-13}{3}$ or $x=5$
Now, $x \neq \frac{-13}{3}$ as it does not satisfy the (i)
$\therefore$ Hence, the roots of the given equation are 4 and 5.

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