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?
 A: Yes , since we have $$3\geq a$$ you can square the inequality with $+$ sign and you will get $$3-a<(3-a)^2$$ so you will get $$0<(3-a)(3-a-1)$$
$$-\sqrt{3-a}<3-a$$ is fulfilled for $$3>a$$
A: You don't have to calculate the roots, nor square them – only use high-school theorems on the sign of quadratic polynomials:


*

*The polynomial $p(x)=x^2 − 2ax + a^2 + a – 3$ has real roots, so its reduced discriminant is non-negative:
$$\Delta'=a^2-(a^2+a-3)=3-a \ge 0,\text{ i.e. } a\le 3.$$

*$3$ has to be outside the interval of the roots $x_1, x_2$: this means $p(3)=a^2-5a+6>0$. As the roots of $a^2-5a+6$ are $2$ and $3$, we must have $a<2\quad\text{or}\quad a>3$.

*From condition 2, we know that either $3< x_1<x_2$ or $x_1<x_2<3$. In other words either  both roots are greater than $3$ or both are less. To ensure that we're in the latter case, it is enough to ensure their arithmetic mean is less than $3$. Now
$$\frac{x_1+x_2}2=a$$
so the condition is $a<3$.


Combining all these conditions, we obtain $\;a\in \color{red}{(-\infty,2)}$.
A: The inequality $\;\underbrace{- \sqrt {3 – a } }_{{-}}\leq \underbrace{3 – a}_{+}\;$ holds trivially.
As $\;3-a\geq 0\;$ and  $\;\sqrt {3 – a } \geq 0,\;$ squaring $\sqrt {3 – a } \geq 3 – a\;$ is legitimate, as stated by @auscrypt.
A: 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 $3-a>1$
A: You can square both sides of an inequality $x<y$ to get $x^2<y^2$, provided $x\geq 0$.
A: Note that you will have different results while squaring both sides of 
$$ \sqrt {3 - a } \lt 3 - a$$
and
$$- \sqrt {3 - a } \lt 3 - a$$
If you take square of both sides by taking this into consideration, it is allowable.
A: It's not always allowable; take $a = 2.99$, then $-\sqrt{3-a} = -0.1 < 0.01 = 3-a$. But squaring both sides gives $0.01<0.0001$, which is false. The issue is that negative signs cause issues when you square, since the absolute value of the left side may be larger than the right.
To solve this problem, note that both roots are real when $a\le 3$, and note that the largest root, $a + \sqrt{3-a}$ is less than $3$. So $a + \sqrt{3-a} < 3$, and so $\sqrt{3-a} < 3-a$. Here, we are allowed to square, because there is no negative sign, which is the key. Then you solve the quadratic, which is straightforward.
A: You can avoid squareing inequality. 
Since $2a =x_1+x_2 <6$ we have $a<3$ and $$a^2+a-3=x_1x_2 < 9$$ we have also $a^2+a-12 =(a+4)(a-3)<0$ so $a\in (-4,3)$. 
A: $y=(x-a)^2+(a-3)$;
$a-3 >0:$ 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:
$a \in (-\infty, 2)$
