SOLUTION: solve b^4 + 2b^2 - 24 = 0 a) -2, -sqrt6, sqrt6, 2 b) -sqrt6, 2, 2i, i sqrt6 c) -2, 2, -i sqrt6, i sqrt6 d) -2i, 2i

SOLUTION: solve b^4 + 2b^2 - 24 = 0 a) -2, -sqrt6, sqrt6, 2 b) -sqrt6, 2, 2i, i sqrt6 c) -2, 2, -i sqrt6, i sqrt6 d) -2i, 2i - sqrt6, sqrt6

Question 296657: solve b^4 + 2b^2 - 24 = 0
a) -2, -sqrt6, sqrt6, 2
b) -sqrt6, 2, 2i, i sqrt6
c) -2, 2, -i sqrt6, i sqrt6
d) -2i, 2i - sqrt6, sqrt6
Answer by richwmiller(17219) (Show Source): You can put this solution on YOUR website!
I'll get you started.
Since all complex roots come in plus minus pairs, answer b is out
To solve the equation, substitute u for b^2
b^4 + 2b^2 - 24 = 0
u^2+2u-24=0
(u-4)*(u+6)=0
now plug b^2 back in for u
(b^2-4)*(b^+6)=0
(b-2)*(b+2)*(b^2+6)=0
solve for b
so b = +-2
answer d is out
b^2=-6
so answer a is out
so it must be answer c
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