<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] $y=(x-a)^2+(a-3)$;
$a-3 >0:$ This parabola does not cut the $x-$axis ($(x-a)^\ge 0)$.
Hence $a-3\le 0$.
Roots:
$y=(x-a)^2 +(a-3) = 0;$
$x_{1,2}=a\pm \sqrt{3-a} <3$.
1)Let $0\le a \lt 3$ ($a=3$ is ruled out)
$\sqrt{3-a} <3-a=\sqrt{3-a}\sqrt{3-a}$.
This is true for $\sqrt{3-a} >1$, or $ 3-a >1$.
Hence $2>a \ge 0$.
2) Let $a <0$;
Then $-|a| +\sqrt{3+|a|} <3$.
$\sqrt{3+|a|} <3+|a|=\sqrt{3+|a|}\sqrt{3+|a|}.$
This is true for $\sqrt{3+|a|} >1$, or $3+|a|>1$.
Hence $a<0$;
Combining:
$a \in (-\infty, 2)$
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