SOLUTION: Find all real and imaginary solution to the equation. b^4+13b^2+36=0

Algebra ->  Square-cubic-other-roots -> SOLUTION: Find all real and imaginary solution to the equation. b^4+13b^2+36=0      Log On


   

Question 122517: Find all real and imaginary solution to the equation.
b^4+13b^2+36=0
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!

Let x=b%5E2, then x%5E2=b%5E4.

Substitute:
b%5E4%2B13b%5E2%2B36=0
x%5E2%2B13x%2B36=0

The quadratic factors:
%28x%2B4%29%28x%2B9%29=0

Therefore
x=-4 or x=-9

But x=b%5E2 so

b%5E2=-4 or b%5E2=-9

Therefore

b=2i
b=-2i
b=3i
b=-3i

Giving us the 4 expected roots for the quartic (degree 4) equation.

Since the roots are all complex, there are no real zeros for this equation,
therefore the graph will not intercept the x-axis anywhere.


graph%28600%2C600%2C-5%2C5%2C-5%2C80%2Cx%5E4%2B13x%5E2%2B36%29