SOLUTION: Pleae help. Find all real and imaginary solutions to each equation. b^4+13b^2+36=0

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Question 179152: Pleae help.
Find all real and imaginary solutions to each equation.
b^4+13b^2+36=0
Answer by monika_p(71) About Me  (Show Source):
You can put this solution on YOUR website!
b^4+13b^2+36=0
substitute b^2=t
t^2+13t+36=0
discriminant: b^2-4ac=13^2-4*1*36=25
t1=-13-sqrt(25)/2*1
t1=-9
t2=-13+sqrt(25)/2*1
t2=-4
Now we have to go back to substitution b^2=t and substitute t1 and t2 in the place of t
for b^2=-9
b=sqrt(-9)
b1=sqrt(9*-1)=sqrt(9*i^2)
b1=3i
for b^2=-4
b=sqrt(-4)
b2=sqrt(4*-1)=sqrt(4*i^2)
b2=2i
answer: b1=3i b2=2i