How do you find the complex roots of #b^2+36=0#? Precalculus Polynomial Functions of Higher Degree Zeros 1 Answer sjc Dec 16, 2016 #b=+-6i# Explanation: rearrange as normal #b^2=-36# #b=sqrt(-36)# #b=+-sqrt(-1)sqrt36# #b=+-6i# Answer link Related questions What is a zero of a function? How do I find the real zeros of a function? How do I find the real zeros of a function on a calculator? What do the zeros of a function represent? What are the zeros of #f(x) = 5x^7 − x + 216#? What are the zeros of #f(x)= −4x^5 + 3#? How many times does #f(x)= 6x^11 - 3x^5 + 2# intersect the x-axis? What are the real zeros of #f(x) = 3x^6 + 1#? How do you find the roots for #4x^4-26x^3+50x^2-52x+84=0#? What are the intercepts for the graphs of the equation #y=(x^2-49)/(7x^4)#? See all questions in Zeros Impact of this question 1382 views around the world You can reuse this answer Creative Commons License