<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] To solve $4x^3+12x^2+15x+5=0$, start by applying the rational root theorem.  In this case, potential rational roots are $\pm \lbrace \frac{1}{4}, \frac{1}{2}, 1, \frac{5}{4}, \frac{5}{2}, 5 \rbrace$.  Trying them out, we see that $x=\frac{-1}{2}$ satsfies the equation.  So $x + \frac{1}{2}$ (or for the sake of avoiding fractions, $2x + 1$) is a factor.
$$4x^3+12x^2+15x+5 = (2x + 1)(ax^2 + bx + c)$$
The quadratic term works out to $2x^2 + 5x + 5$.  By the quadratic formula, $x = \frac{-5 \pm i\sqrt{15}}{4}$.
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