document.write( "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? \n" ); document.write( "

Algebra.Com's Answer #171580 by jsmallt9(3758) $\"\"$ $\"About$

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The equation described would be of the form:
\n" ); document.write( " $\"ax%5E4+%2B+cx%5E2+=+0\"$
\n" ); document.write( "From this equation we can factor out x^2:
\n" ); document.write( " $\"x%5E2%28ax%5E2+%2B+c%29+=+0\"$
\n" ); document.write( "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.

\n" ); document.write( "So the 4 roots are: 0 (twice), $\"3%2Bi%2Asqrt%287%29\"$ and $\"3-i%2Asqrt%287%29\"$ \n" ); document.write( "

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