<s>[INST] 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?
 [/INST] $$(x+4)(x-4) + \left(\frac{3x}{x+3}\right)^2 = 0$$
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 : $x=-4\pm\iota 2\sqrt{2}$
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