Solve.
-$z ^ { 4 } - 11 z ^ { 2 } + 30 = 0$
A) $\sqrt { 5 } , \sqrt { 6 }$
B) $\sqrt { - 5 } , \sqrt { - 6 }$
C) $\sqrt { - 5 } , \sqrt { 6 }$
D) $\pm \sqrt { 5 } , \pm \sqrt { 6 }$

Question

Quizplus · Accepted Answer

This equation can be solved by substituting $z^2 = x$, turning it into a quadratic equation $x^2 - 11x + 30 = 0$. Factoring this gives $(x - 5)(x - 6) = 0$, so $x = 5$ or $x = 6$. Reverting back to $z$, we get $z^2 = 5$ and $z^2 = 6$, which have the solutions $\pm\sqrt{5}$ and $\pm\sqrt{6}$, respectively.

Solve - z4−11z2+30=0z ^ { 4 } - 11 z ^ { 2 } + 30 = 0z4−11z2+30=0

Solve
- $z ^ { 4 } - 11 z ^ { 2 } + 30 = 0$