Solve the equation. $- z ^ { 4 } + 26 z ^ { 2 } - 25 = 0$
A) $z = 1 , z = - 1 , z = 4 , z = - 4$
B) $z = 1 , z = 5$
C) $z = 1 , z = 25$
D) $z = 1 , z = - 1 , z = 2 , z = - 2$
E) $z = 1 , z = - 1 , z = 5 , z = - 5$

Question

Quizplus · Accepted Answer

This equation can be solved by substituting $y = z^2$, resulting in a quadratic equation $y^2 - 26y + 25 = 0$. Solving this gives $y = 1$ and $y = 25$, which when substituted back for $z^2$ gives the solutions $z = \pm 1$ and $z = \pm 5$.

Solve the Equation −z4+26z2−25=0- z ^ { 4 } + 26 z ^ { 2 } - 25 = 0−z4+26z2−25=0

Solve the Equation $- z ^ { 4 } + 26 z ^ { 2 } - 25 = 0$