Solve the equation $$z^2+\frac{9z^2}{(3+z)^2}=-5$$
PS.: The expanded form of a 4 degree polynomial is $$z^4+6z^3+23z^2+30z+45=0$$
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2 Answers
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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.
answered Apr 18, 2019 at 14:00
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2$\begingroup$ 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. $\endgroup$ Apr 18, 2019 at 14:04
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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?
answered Jun 10, 2019 at 3:40