Solve the inequality \[\sqrt{x+2}-\sqrt{x-1}>\sqrt{2x-3}\] for real values of \(x\) and of the square roots.

Square both sides of the inequality (this is permissible as the left hand side is positive, since \(\sqrt{x+2}>\sqrt{x-1}\) for real values of the square root): \[\begin{align*} && x+2 -2\sqrt{(x-1)(x+2)} + x-1 &> 2x-3&&\quad \\ \iff&& 2x+1 -2\sqrt{(x-1)(x+2)} &> 2x-3 \\ \iff&& \sqrt{(x-1)(x+2)} &< 2. \end{align*}\]

Squaring again gives \[(x-1)(x+2)<4\] which rearranges to \[x^2+x-6<0.\]

By factorising the quadratic expression as \((x+3)(x-2)\) we can see that the roots are \(x=-3\) and \(2\).

A quick sketch of the quadratic \(y=x^2+x-6\) shows that the inequality \(x^2+x-6<0\) is satisfied when

\[\begin{equation} -3< x <2. \label{eq:1} \end{equation}\] However, considering the original inequality we see that \[\begin{align*} \sqrt{x+2} &\quad\text{is real for} \quad x\geq -2 \\ \sqrt{x-1} &\quad\text{is real for} \quad x\geq 1 \\ \sqrt{2x-3} &\quad\text{is real for} \quad x\geq \frac{3}{2} \end{align*}\]

We must therefore restrict the inequality \(\eqref{eq:1}\) to \[\frac{3}{2}\leq x < 2.\]

Do we need to check that our solution works in the original inequality, since we squared the inequality along the way?

It turns out that we are fine on this occasion.

If we have an inequality \(\sqrt{\vphantom{b}a}>\sqrt{b}\) and square it to get \(a>b\) (where both \(a\) and \(b\) are non-negative, and we have taken the positive square roots), then we can square root to get back to \(\sqrt{\vphantom{b}a}>\sqrt{b}\).

In other words, \[\sqrt{\vphantom{b}a}>\sqrt{b}\quad\iff\quad a>b.\]

However, it’s not true in general that \(a^2>b^2\) if and only if \(a>b\); if \(a=1\) and \(b=-2\), for example, then \(a>b\) but \(a^2<b^2\).

In our case, we relied upon the extra knowledge that both of the square roots were non-negative.