Can $ x^2+\frac{x^2}{(x+1)^2}=3 $ be solved by 'completing square method'?

Question

On using the quartic formula, there are two zeroes of this equation

$$ x^2+\frac{x^2}{(x+1)^2}=3 $$

So, I was wondering if this equation could be solved like a quadratic equation.

Is it possible. If yes, how?

Answer 1

We may notice that the golden ratio $\varphi=\frac{1+\sqrt{5}}{2}$ is a solution of the given equation, since $\varphi^2=\varphi+1$ implies $$ \varphi^2+\frac{\varphi^2}{(\varphi+1)^2} = \varphi^2+\varphi^{-2} = L_2 = 3. $$ In particular, the algebraic conjugate of $\varphi$, given by $\bar{\varphi}=\frac{1-\sqrt{5}}{2}$, is also a solution of the given equation, and the polynomial $x^2-x-1$ (i.e. the minimal polynomial of $\varphi$) is a divisor of $$ x^2(x+1)^2+x^2-3(x+1)^2.$$ By performing a polynomial division, we get that the other solutions are given by $x^2+3x+3=0$, i.e. by $x=\frac{-3\pm i\sqrt{3}}{2}$.

Answer 2

Write $ x^2+\frac{x^2}{(x+1)^2}=3 $ as $$x^2(x+1)^2+x^2=3(x+1)^2$$ or $$x^2(x+1)^2+2x(x+1)+1= 4(x+1)^2$$ which allows the completion of square $$[x(x+1)+1]^2=[2(x+1)]^2$$ Then, factorize with $a^2-b^2=(a+b)(a-b)$ $$(x^2-x-1)(x^2+3x+3)=0$$