Solve: $\sqrt{x^2-4}=x-2$

Question

Solve for x

$$\sqrt{x^2-4}=x-2$$

My try:

$$\sqrt{(x-2)(x+2)}=x-2$$

$$\sqrt{x+2}=\sqrt{x-2}$$

$$x+2=x-2$$

$$0x=4$$

This is not correct $x={4\over 0}$

How else can I solve this equation?

Answer 1

You lost a solution $x=2$ when you divide it by $\sqrt{x-2}$.

You can divide the equation with something only if you are sure it is not $0$.

Answer 2

We can square both side and then divide by $(x-2)$ with the extra condition $x\neq 2$

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

then check directly the solution $x=2$ in the original equation.

Answer 3

$$\sqrt{x^2-4}=x-2$$

squaring both side,

$${(x-2)(x+2)}=(x-2)^2$$

$$(x-2)((x+2)-(x-2))=0$$

$$(x-2)(4)=0$$

$$x = 2$$

In Your approach, you lost this root $x=2$ when you divided by $\sqrt{x-2}$

Answer 4

Original equation: $$\sqrt{x^2-4}=x-2$$ Square on both sides: $$x^2-4=x^2-4x+4$$ "Move things": $$-4x=-8\\4x=8\\x=\frac{8}{4}=2$$

Now, let's test the solution: $$\sqrt{2^2-4}\stackrel{?}{=}2-2\\\sqrt{0}\stackrel{?}{=}0\\0=0$$ It does work, so $x=2$