SOLUTION: b^4+13b^2+36=0

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Question 80885This question is from textbook algebra
: b^4+13b^2+36=0 This question is from textbook algebra

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 Let's temporarily swap variables: b%5E2+=+x so then b%5E4+=+x%5E2 Now rewrite the equation:
x%5E2%2B13x%2B36+=+0 Solve for x by factoring.
%28x%2B4%29%28x%2B9%29+=+0 Apply the zero products principle:
x%2B4+=+0 and/or x%2B9+=+0
If x%2B4+=+0 then x+=+-4
If x%2B9+=+0 then x+=+-9
Now replace the x with b%5E2
For x+=+-4 we get:
b%5E2+=+-4 Take the square root of both sides.
b+=+sqrt%28-4%29 or b+=+-sqrt%28-4%29
b+=+2sqrt%28-1%29 or b+=+-2sqrt%28-1%29
b+=+2i or b+=+-2i
For x+=+-9 we get:
b%5E2+=+-9 Take the square root of both sides.
b+=+sqrt%28-9%29 or b+=+-sqrt%28-9%29
b+=+3sqrt%28-1%29 or b+=+-3sqrt%28-1%29
b+=+3i or b+=+-3i
So the four solutions are: i+=+sqrt%28-1%29
b+=+2i
b+=+-2i
b+=+3i
b+=+-3i