How do you find all the real and complex roots of #x^4-12x^2+27=0#? Precalculus Complex Zeros Complex Conjugate Zeros 1 Answer GiĆ³ Jan 9, 2016 I found four roots: Explanation: Try this: Answer link Related questions What is a complex conjugate? How do I find a complex conjugate? What is the conjugate zeros theorem? How do I use the conjugate zeros theorem? What is the conjugate pair theorem? How do I find the complex conjugate of #10+6i#? How do I find the complex conjugate of #14+12i#? What is the complex conjugate for the number #7-3i#? What is the complex conjugate of #3i+4#? What is the complex conjugate of #a-bi#? See all questions in Complex Conjugate Zeros Impact of this question 1979 views around the world You can reuse this answer Creative Commons License