SOLUTION: A quartic equation with integral coefficients has no cubic term and no constand term. If one root is 3-i the square root of 7, what are the other roots?

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Question 231934: A quartic equation with integral coefficients has no cubic term and no constand term. If one root is 3-i the square root of 7, what are the other roots?
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
The equation described would be of the form:
ax%5E4+%2B+cx%5E2+=+0
From this equation we can factor out x^2:
x%5E2%28ax%5E2+%2B+c%29+=+0
From this we can see that x=0 is a double root (aka root of multiplicity 2). And if 3-i%2Asqrt%287%29 is a root, then its conjugate. 3%2Bi%2Asqrt%287%29, will also be a root.

So the 4 roots are: 0 (twice), 3%2Bi%2Asqrt%287%29 and 3-i%2Asqrt%287%29