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?
$\endgroup$ 14 Answers
$\begingroup$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 $0$.
$\endgroup$ $\begingroup$$$\sqrt{x^2-4}=x-2$$
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 $\sqrt{x-2}$
$\endgroup$ $\begingroup$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 $x=2$
$\endgroup$ 2 $\begingroup$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.
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