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

Question

How do I solve this equation I found in my textbook:

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

This is what I tried:

$\sqrt{x^2-16} - (x-4) = \sqrt{x^2-5x+4}$ $\mapsto \sqrt{(x+4)(x-4)} - (x-4) = \sqrt{(x-1)(x-4)}$

Dividing both sides by $\sqrt{x-4}$

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

Squaring both Sides

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

Squaring both sides

$ \mapsto \frac{x^2 + 2x + 1}{4} = x^2 - 16$ $\mapsto 3x^2 -2x - 65 = 0$

Solving the quadratic equation

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

Checking in the initial equation we can see that $5$ is a valid root. But the second value that is given in the key to the book is $4$. How do I obtain $4$?

Answer 1

The domain gives $x\geq4$ or $x\leq-4$.

1. $x\leq-4$.

We need to solve $$\sqrt{(4-x)(-4-x)}+\left(\sqrt{4-x}\right)^2=\sqrt{(4-x)(1-x)}$$ or $$\sqrt{-4-x}+\sqrt{4-x}=\sqrt{1-x}$$ or $$-4-x+4-x+2\sqrt{x^2-16}=1-x$$ or $$2\sqrt{x^2-16}=x+1,$$ which has no real roots.

2. $x\geq4.$

We have $$\sqrt{(x-4)(x+4)}=\left(\sqrt{x-4}\right)^2+\sqrt{(x-4)(x-1)},$$ which gives $$x=4$$ or $$\sqrt{x+4}=\sqrt{x-4}+\sqrt{x-1},$$ which is $$x+4=x-4+x-1+2\sqrt{x^2-5x+4}$$ or $$2\sqrt{x^2-5x+4}=9-x,$$ which gives $$4\leq x\leq9$$ and $$4(x^2-5x+4)=(9-x)^2$$ or $$3x^2-2x-65=0,$$ which gives $$x=5$$ and the answer: $$\{4,5\}.$$

Answer 2

At the stage where you divided by $\sqrt{x-4}$, you forgot that $\sqrt{x-4}$ could be $0$, which gives the solution $x=4$. You can only divide by $\sqrt{x-4}$ if $x-4\neq 0$.

Answer 3

I think it should be like this when the equation gets divided by $\sqrt{x-4}$:

$(\sqrt{x-4})(\sqrt{x+4} - \sqrt{x-4}) - \sqrt{(x-4)(x-1)} = 0$ $\mapsto (\sqrt{x-4})(\sqrt{x+4} - \sqrt{x-4} - \sqrt{x-1}) = 0$

$\therefore$ Either $\sqrt{x-4} = 0$ OR $(\sqrt{x+4} - \sqrt{x-4} - \sqrt{x-1}) = 0$

Answer 4

HINT.-Your equation is the same as $$\sqrt{(x-4)(x+4)}-\sqrt{(x-4)(x-4)}=\sqrt{(x-4)(x-1)}$$ which gives besides of $x-4=0$ the equation $$\sqrt{x+4}-\sqrt{x-4}=\sqrt{x-1}\iff4(x^2-16)=(x+1)^2$$