Solve the equation.
-$$2 a ^ { 4 } + 1 = 7 a ^ { 2 }$$
A) $\left\{ - \frac { \sqrt { 7 + \sqrt { 41 } } } { 2 } , - \frac { \sqrt { 7 - \sqrt { 41 } } } { 2 } , \frac { \sqrt { 7 - \sqrt { 41 } } } { 2 } , \frac { \sqrt { 7 + \sqrt { 41 } } } { 2 } \right\}$
B) $\left\{ \frac { - 7 - \sqrt { 41 } } { 4 } , \frac { - 7 + \sqrt { 41 } } { 4 } \right\}$
C) $\left\{ \frac { 7 - \sqrt { 41 } } { 4 } , \frac { 7 + \sqrt { 41 } } { 4 } \right\}$
D) $\left\{ - \frac { \sqrt { 7 + \sqrt { 41 } } } { 2 } i , - \frac { \sqrt { 7 - \sqrt { 41 } } } { 2 } i , \frac { \sqrt { 7 - \sqrt { 41 } } } { 2 } i , \frac { \sqrt { 7 + \sqrt { 41 } } } { 2 } i \right\}$

Question

Quizplus · Accepted Answer

The equation can be rewritten as a quadratic in terms of $a^2$: $2a^4 - 7a^2 + 1 = 0$. Solving this quadratic equation for $a^2$ yields two solutions, which when square-rooted give four real solutions for $a$, matching the roots in option A.

Solve the Equation 2a4+1=7a22 a ^ { 4 } + 1 = 7 a ^ { 2 }2a4+1=7a2

Solve the Equation $2 a ^ { 4 } + 1 = 7 a ^ { 2 }$