Solving $z^2+\frac{9z^2}{(3+z)^2}=-5$

Solve the equation $$z^2+\frac{9z^2}{(3+z)^2}=-5$$

PS.: The expanded form of a 4 degree polynomial is $$z^4+6z^3+23z^2+30z+45=0$$

The equation is equivalent to $$z^4 + 6z^3 + 23z^2 + 30z + 45=(z^2 + 5z + 15)(z^2 + z + 3)=0$$ for all $$z\neq -3$$. We can solve the quadratic equations, and hence also the degree $$4$$ equation. There are no real solutions. Of course, this is already clear, since $$z^2+a^2=-5$$ is impossible over the real numbers.

• Thanks! But how did you factored? Apr 18, 2019 at 14:02
• Just by multiplying out $(x^2+ax+b)(x^2+cx+d)$ and comparing coefficients. This goes fast and easy. If you only consider real solutions, we don't have to factorize at all. Apr 18, 2019 at 14:04

We have $$z^2-\frac{6z^2}{3+z}+\frac{9z^2}{(3+z)^2}+\frac{6z^2}{3+z}+5=0$$ or $$\left(z-\frac{3z}{3+z}\right)^2+\frac{6z^2}{3+z}+5=0$$ or $$\left(\frac{z^2}{3+z}\right)^2+\frac{6z^2}{3+z}+5=0$$ or $$\left(\frac{z^2}{3+z}+1\right)\left(\frac{z^2}{3+z}+5\right)=0.$$ Can you end it now?