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.
 A: Simplify as
$$4x^2[(x+1)^2+1]=5(x+1)^2 \implies 4x^4+8x^3+3x^2-10x-5=0$$
Denoting LHS as $f(x)$, note that $f(1)=0 ~\& ~f(-1/2)=0.$ This means that by remainder theorem $$f(x) ~\text{is divisible by}~ (2x+1)(x-1)=2x^2-x-1=g(x)$$
dividing $f(x)$ by $g(x)$ we see that $$f(x)=(2x^2-x-1)(2x^2+5x+5)$$
By solving $2x^2+5x+5=0$ we get $$x=\frac{-5\pm \sqrt{-15}}{4}=\frac{-5\pm i\sqrt{15}}{4}.$$
Finally, the equation has 4 roots $$x=-1/2,1,\frac{-5\pm i\sqrt{15}}{4}.$$
A: $x^2 + \left(\frac{x}{x+1}\right)^2 = \frac{5}{4}$
Let $y=-x\qquad ⇒ y^2 + \left(\frac{y}{1-y}\right)^2 = \frac{5}{4}$
Let $z=\frac{x}{x+1}\qquad ⇒ z^2 + \left(\frac{z}{1-z}\right)^2 = \frac{5}{4}$
By inspection, $(x=1)$ is a solution.
$(x=1) → (y=-1) → (z=-1)$
Solve $z=-1$, back for x, we have $\;\left(x=-\frac{1}{2}\right)$
Original formula can be rewritten as quartic polynomial.
With 2 known roots, it reduced to simple quadratics.
A: 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, $x = \frac{-5 \pm i\sqrt{15}}{4}$.
A: 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
$\displaystyle s=-\frac{1}{2}\qquad ⇒ x = \frac{-5 ± i \sqrt{15}}{4}$
$\displaystyle s=+\frac{5}{2}\qquad ⇒ x = \frac{1 ± 3}{4}$
