I've a small problem solving this inequality

Question

$\sqrt{9 - x^2} \geq \sqrt{3}*x$

I used the standard method (meaning, it involves two systems of inequalities).

At the end I've got (on the number line): $x>-3$, $x<3$, $-1.5<x<1.5$, $x>0$, and $x < 0$.

I don't understand the reason why in the solution set I have to include "$-3<x<-3/2$" because this set isn't satisfied by all the other set of solutions. (because of union, it's not the case of intersection).

Answer 1

Define $$f(x)=\sqrt{9-x^2},\quad g(x)=\sqrt3x,\quad h(x)=f(x)-g(x).$$ We want to find where $f(x)\geq g(x)\implies h(x)\geq0$. We can first find where $h(x)=0$, then determine whether $h(x)$ is positive or negative on the intervals that have been created by the above solutions.

\begin{align} h(x)&=0\\ \sqrt{9-x^2}-\sqrt3x&=0\\ \sqrt{9-x^2}&=\sqrt3x\\ 9-x^2&=3x^2\\ 0&=4x^2-9\\ x&=\pm\frac32 \end{align}

$h(x)$ is only defined on $[-3,3]$, so from that we can create the following table of values: \begin{array}{|c|c|c|c|c|} \hline\text{Interval}&h(x)\\ \hline x=-3 & 3\sqrt3\\ \hline-3<x<-\frac32 & +\\ \hline x=-\frac32 & 0\\ \hline -\frac32<x<\frac32 & +\\ \hline x=\frac32 & 0\\ \hline \frac32<x<3 & -\\ \hline x=3 & -3\sqrt3\\ \hline \end{array}

Therefore, the solution is $\boxed{x\in\left[-3,\frac32\right]}$.

Answer 2

For real $x,\sqrt{9-x^2}\ge0$ and $x^2\le9\iff-3\le x\le3$

So, if $-3\le x<0, \sqrt{9-x^2}>\sqrt3x$

If $x\ge0,$ $$ \sqrt{9-x^2}\ge\sqrt3x\iff9-x^2\ge(\sqrt3x)^2\iff x^2\le\dfrac94$$

Now if for $a\ge0,x^2\le a^2,-a\le x\le a$

and if $x\ge0, 0\le x\le a$