sum of squares of all real values of x satisfying given equation

Question

If $x \in \mathbb{R}$ such that $x^2+\frac{9x^2}{(x+3)^2}=27$, then sum of squares of all real values of $x$ satisfying given equation is _________.

Answer 1

It's $$x^2-\frac{6x^2}{x+3}+\frac{9x^2}{(x+3)^2}+\frac{6x^2}{x+3}=27$$ or $$\left(x-\frac{3x}{x+3}\right)^2+\frac{6x^2}{x+3}=27$$ or $$\left(\frac{x^2}{x+3}\right)^2+\frac{6x^2}{x+3}-27=0,$$ which gives $$\frac{x^2}{x+3}=-9$$ or $$\frac{x^2}{x+3}=3.$$ The first equation gives $$x^2+9x+27=0$$ or $$(x+4.5)^2+6.75=0,$$ which is impossible.

The second case for you (use the Viete's theorem).

I got $x_1^2+x_2^2=27$.