Find $x$, given: $x^2 + \frac{9x^2}{(x+3)^2} = 16$ Here's an equation:
$$x^2 + \frac{9x^2}{(x+3)^2} = 16$$
First, I subtracted 16 from both sides and factored $x^2$ so I would get a quadratic equation, but with no success. Also, I can see that the equation can be rewritten as:
$$(x+4)(x-4) + \left(\frac{3x}{x+3}\right)^2 = 0$$
But I can't see how can I use that information.
What should I do?
 A: You can rewrite it as:
\begin{align*}
\left(\frac{x}{4}\right)^2 + \left(\frac{3x}{4(x+3)}\right)^2 & = 1.
\end{align*}
Let $\frac{x}{4}=\cos A$, then $\frac{3x}{4(x+3)}=\sin A$. Using $x=4 \cos A$ in the expression for $\sin A$, we get
\begin{align*}
12 (\cos A -\sin A) & = 8 \sin 2A\\
9(\cos A -\sin A)^2 & = 4 \sin ^22A\\
9(1-\sin 2A)&=4 \sin^2 2A.
\end{align*}
Now you have a quadratic in $\sin 2A$. Can you proceed from here?
Comment: (added)
This quadratic gives $\sin2A= -3$ or $\sin 2A=\frac{3}{4}$. The former cannot happen (for real values of $x$). The latter upon substituting back in terms of $x$ gives two real solutions $x=1\pm \sqrt{7}$. 
A: Hint:
Multiply by $(x+3)^2$,
$$x^2(x+3)^2+9x^2=16(x+3)^2$$
or
$$x^4+6x^3+2x^2-96x-144=0.$$
Using a solver, the roots are irrational, but there is a rational factorization in two quadratic polynomials.
A: $$(x+4)(x-4) + \left(\frac{3x}{x+3}\right)^2 = 0$$
multiply by $(x+3)^2$
$$(x+4)(x-4)(x+3)^2 + 9x^2 = 0$$
$$x^4+6x^3+2x^2-96x-144=0$$
$$(x^2-2x-6)(x^2+8x+24)=0$$
real solutions :  $x=1\pm \sqrt{7}$
complex solutions : $x=-4\pm\iota 2\sqrt{2}$
