<s>[INST] How to tackle this squaring of inequality problem 
If the roots of quadratic equation     $$x^2 − 2ax + a^2 + a – 3 = 0$$ 
  are real and less than $3$, find the range of $a$.

The roots are   $a \pm \sqrt {3 – a}$
For the roots to be real, we must have  a < 3.
Also, for the roots to be less than 3, we must have $\pm \sqrt {3 – a } \lt 3 – a $
If squaring both sides is allowable, I will get $(a – 2)(a – 3) > 0$. Then the problem is solved.
The question is:- how to convince others that the squaring of both sides of $\pm \sqrt {3 - a } \lt 3 - a $ is allowable?
 [/INST] Since $x_1,x_2<3$ we have $3-x_i>0$ so their product is positive:$$ 0<9-3(x_1+x_2)+x_1x_2$$
thus $$0<9-6a+a^2+a-3 = a^2-5a+6=(a-3)(a-2)$$
So $a\in (-\infty ,2)\cup (3,\infty)$. But since the discriminat must be $\geq 0$ we get $a\leq 3$ so we have $a\in (-\infty ,2)$
Now let us prove that all $a<2$ are good.
We have $$x_1= a-\sqrt{3-a}\leq a<3$$ We are left if $x_2= a+\sqrt{3-a}$ is smaller than $3$ if $a<2$ i.e. $$\sqrt{3-a} <3-a$$ which is true since $3-a>1$
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