<s>[INST] Solve $\sqrt {x^2-3}=x-3$ in $\mathbb R$ 
Solve $\sqrt {x^2-3}=x-3$ in $\mathbb R$

My attempt:
$|x^2-3|=(x-3)^2$
So $-(x^2-3)=(x-3)^2$ or $(x^2-3)=(x-3)^2$
If  $-(x^2-3)=(x-3)^2=x^2+9-6x$ 
So no solutions in $\mathbb R$
And if $(x^2-3)=(x-3)^2$
So $x^2-3=x^2+9-6x$
Now, can I delete $x^2$ with $x^2$ ? Like this 
$x^2-x^2-3-9+6x=0$
$6x=12$
$x=2$
But $f(2)$ isn’t equal to $0$?
 [/INST] What you solved is correct but you must notice that $$\sqrt {x^2-3}\ge 0\Rightarrow x-3\ge 0\Rightarrow x\ge 3$$
And hence I'd you get any solution less than $3$ then you have to reject it.
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