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 $0$?
 A: $$\sqrt {x^2-3}=x-3$$
Since, by definition, $\sqrt {x^2-3} \ge 0$, we must have $x-3\ge 0$. That is we must have $x \ge 3$.
If $x \ge 3$, then $x^2-3 \ge 6$. So there will be no complications caused by squaring both sides of the equation.
\begin{align}
   x^2-3 &= x^2-6x + 9 \\
   6x &= 12 \\
   x &= 2
\end{align}
Since we need $x \ge 3$, there is no solution.
A: 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.
A: When you solve an equation by squaring both sides, it's possible that extraneous solutions are introduced.  An extraneous solution is a value that you get at the end (like $x=2$ here) that doesn't actually solve the original equation.  This is exactly what happened here.  Since $x=2$ is the only value you got, and since it doesn't solve the original equation, that means the equation has no real solutions.
This can also be verified by graphing $y = \sqrt{x^2-3}$ and $y = x-3$ on the same axes and noticing that they never intersect.
Note also that $\left(\sqrt{x^2-3}\right)^2$ is just $x^2 - 3$, not $|x^2-3|$.  So considering $-(x^2 - 3) = (x-3)^2$ is technically incorrect, but you got lucky here in that it also has no real solutions.  Still, you should remove it completely from your work.
A: From
$$\sqrt {x^2-3}=x-3$$
what we need stricktly as condition is that $x^2-3\ge 0$, then we can square both sides
$$\sqrt {x^2-3}=x-3\iff x^2-3=(x-3)^2$$
and for $x\neq 3$, noting that $x=3$ is not a solution, we obtain
$$\iff x^2-3=x^2-6x+9 \iff 6x=12\iff x=2$$
but since $x=2$ doesn't satisfy the equation it is to reject.
This is a legit way to solve, the extra solution come in from squaring both sides but we can exclude it at the end by checking directly on the original equation. 
As an alternative we can observe that since LHS is $\ge 0$ also RHS must be then $x-3\ge0 $ is required by the original equation.
A: I was taught to always find the domain of possible solutions first. We have 
\begin{cases}
x-3 \ge 0, \\
x^2-3 \ge0 
\end{cases} or \begin{cases}
x \ge 3, \\
-\sqrt{3} \le x \le \sqrt{3}
\end{cases}
which has no solutions 
