SOLUTION: b^4 + 13b^2 + 36 = 0 find all real and imaginary solutions to the equation

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Question 192279This question is from textbook Elementary and intermediate algebra
: b^4 + 13b^2 + 36 = 0 find all real and imaginary solutions to the equation This question is from textbook Elementary and intermediate algebra

Found 2 solutions by Alan3354, Earlsdon:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
b^4 + 13b^2 + 36 = 0 find all real and imaginary solutions to the equation
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(b^2+4)*(b^2+9) = 0
b = +2i, -2i, +3i, -3i

Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Solve for b:
b%5E4%2B13b%5E2%2B36+=+0 Rewrite this as a quadratic equation:
%28b%5E2%29%5E2%2B13%28b%5E2%29+%2B+36+=+0 Factor:
%28b%5E2%2B4%29%28b%5E2%2B9%29+=+0 Apply the zero product rule:
b%5E2%2B4+=+0 or b%5E2%2B9+=+0 so...
b%5E2+=+-4 or b%5E2+=+-9
b%5E2+=+-4 Take the square root of both sides.
b+=+sqrt%28-4%29 or b+=+-sqrt%28-4%29 these can be written as:
b+=+2i or b+=+-2i where i+=+sqrt%28-1%29
b%5E2+=+-9 Take the square root of both sides.
b+=+sqrt%28-9%29 or b+=+-sqrt%28-9%29 these can be written as:
b+=+3i or b+=+-3i
The roots are:
b+=+2i
b+=+-2i
b+=+3i
b+=+-3i