Question 289055: Solve b^4+b^2-20=0
thanks
...by any chance, is this answer correct: -2,2,-i radical 5, i radical 5?
These are my steps...please tell me if I found out the correct answer, and if I didn't, please show me the correct answer....thanks
b^2=x
b^4+b^2-20=0
x^2+x-20=0
x^2+5x-4x-20=0
x(x+5)-4(x+5)=0
(x+5)(x-4)=0
{b^2+5}{b^2-4}=0
(b^2+5)(b+2)(b-2)=0
b=2,-2
b^2=(-1)(5)
b= radical (-1)(5)
b= +/- i radical 5
b^2+b( i sqrt 5)- b( i sqrt 5)- (i^2)(sqrt 5^2)
b^2-(-1)(5)
i^2=-1
=b^2+5
So, is it correct that the answers are: -2,2,-i radical 5, i radical 5?
Answer by nerdybill(7384)   (Show Source):
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