<s>[INST] Solving the equation $x^2+(\frac{x}{x+1})^2=\frac54$ 
Solve the equation$$x^2+\left(\frac{x}{x+1}\right)^2=\frac54$$

I noticed that for $0< x$, both $x^2$ and $\left(\dfrac x{x+1}\right)^2$ are increasing functions so their sum is also increasing and it has only one root which is $x=1$ (by inspection). But I'm not sure how to find the root for negative values of $x$.
By multiplying both sides by $(x+1)^2$ we get,
$$x^2(x^2+2x+1)+x^2-\frac54(x^2+2x+1)=0$$
$$4x^4+8x^3+3x^2-10x-5=0 \implies (x-1)(4x^3+12x^2+15x+5)=0$$
But I can't factor the third degree polynomial.
 [/INST] Let $t=x+1$
$\displaystyle (t-1)^2 + \left( \frac{t-1}{t} \right)^2 = \frac{5}{4}$
$(t^2 + 2 + \frac{1}{t^2}) - 2\left(t + \frac{1}{t} \right) = \frac{5}{4}$
Let $s = \left(t + \frac{1}{t} \right)$
$s^2 - 2s - \frac{5}{4} = (s+\frac{1}{2})(s-\frac{5}{2}) = 0$
We reduced solving quartic for x to quadratic for s, then quadratic for t
$\displaystyle s=-\frac{1}{2}\qquad ⇒ x = \frac{-5 ± i \sqrt{15}}{4}$
$\displaystyle s=+\frac{5}{2}\qquad ⇒ x = \frac{1 ± 3}{4}$
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