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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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.
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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. $\endgroup$ – Dietrich Burde Apr 18 at 14:04
