<s>[INST] Find the range of $\frac{\sqrt{(x-1)(x+3)}}{x+2}$ 
Find the range of $\frac{\sqrt{(x-1)(x+3)}}{x+2}$

My Attempt:$$y^2=\frac{x^2+2x-3}{x^2+4x+4}=z\\\implies  x^2(z-1)+x(4z-2)+4z+3=0$$
Discriminant greater than equal to zero, so, $$(4z-2)^2-4(z-1)(4z+3)\ge0\\\implies z\le\frac43\\\implies -\frac2{\sqrt3}\le y\le\frac2{\sqrt3}$$
But the answer given is $[-\frac2{\sqrt3},1]$
What am I missing?
 [/INST] The problem is that squaring an equation may introduce extraneous solutions. You either have to check each solution (by substituting it into the original equation), or you use equivalent equations throughout the solution process which ensures extraneous solutions won't occur. Taking the second approach, your equation $y=\sqrt{(x-1)(x+3)}/(x+2)$ is equivalent to the system
$$\left\{
\begin{array}{l}
(y^2-1)x^2+2(2y^2-1)x+4y^2+3=0\\
y(x+2)\ge 0\\
x+2\ne 0
\end{array}
\right. $$
The range of $f$ is those $y$ for which the system has a solution. For $y^2-1=0\Leftrightarrow y=\pm1$, the equation is $2x+7=0\Leftrightarrow x=-7/2$. With $y=1$ the first inequality isn't satisfied while $y=-1$ is ok. So $y=-1$ is in the range of $f$ and $y=1$ isn't. For $y^2-1\ne 0$, the quadratic formula gives
$$x=\frac{1-2y^2\pm\sqrt{4-3y^2}}{y^2-1},\quad\quad
y\in\left[-\frac{2}{\sqrt 3},\frac{2}{\sqrt 3}\right]\setminus\{-1,1\} $$
and now you have to solve the inequalities $y(x+2)\ge 0,x+2\ne 0$ with each of the two roots for $x$. That is, you have to solve
$$\frac{y\left(-1-\sqrt{4-3y^2}\right)}{y^2-1}\ge0\\-1-\sqrt{4-3y^2}\ne 0 $$
and
$$\frac{y\left(-1+\sqrt{4-3y^2}\right)}{y^2-1}\ge0\\-1+\sqrt{4-3y^2}\ne 0 $$
The solutions to these are, respectively, $y\in\left[-\frac{2}{\sqrt{3}},-1\right)\cup\left[0,1\right)$ and $y\in\left[-\frac{2}{\sqrt{3}},-1\right)\cup\left(-1,0\right]$. Combining the two sets and also adding $y=-1$ (but not $y=1$), we conclude that the range of $f$ is
$$\left[-\frac{2}{\sqrt{3}},1\right)$$
Edit: For the second pair of inequalities, you can use
$$-1+\sqrt{4-3y^2}
=\frac{\left(-1+\sqrt{4-3y^2}\right)\left(1+\sqrt{4-3y^2}\right)}{1+\sqrt{4-3y^2}}
=\frac{-3(y^2-1)}{1+\sqrt{4-3y^2}} $$
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