SOLUTION: Please help me solve {{{b^4 + b^2 - 20 = 0}}}

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Question 6533: Please help me solve b%5E4+%2B+b%5E2+-+20+=+0
Found 2 solutions by longjonsilver, Shin123:
Answer by longjonsilver(2297) About Me  (Show Source):
You can put this solution on YOUR website!
this is just a quadratic...let a=b%5E2, so the equation becomes

a%5E2+%2B+a+-+20+=+0

so, now factorise to (a+5)(a-4) = 0
--> a = -5 or a = 4

so b%5E2+=+4 or b%5E2+=+-5

so b = +2 or -2 or b = +sqrt%285%29i or -sqrt%285%29i

jon.

Answer by Shin123(626) About Me  (Show Source):
You can put this solution on YOUR website!
Let a=b2. Then the equation becomes a2+a-20=0. Factoring leaves %28a%2B5%29%28a-4%29=0 So a=-5 or a=4. So we have b%5E2=-5 and b%5E2=4 So the real solutions are b=2 and b=-2. (Note:The imaginary solutions are b=i%2Aroot%282%2C5%29 and b=-i%2Aroot%282%2C5%29).