Find the quadratic equation from given relatioshipThe quadratic equation whose roots are a and b wherea2+b2=5 and 3(a5+b5)=11(a3+b3)

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Find the quadratic equation from given relatioshipThe quadratic equation whose roots are a and b wherea2+b2=5 and 3(a5+b5)=11(a3+b3)

resacarno4u · Accepted Answer

There is no single answer here. It turns out there are three quadratics with real coefficients and two more with complex coefficients, the roots of which satisfy the given conditions.The quadratic equation with roots a and b is&amp;nbsp;x2−sx+p, where&amp;nbsp;s=a+b&amp;nbsp;and&amp;nbsp;p=ab. Some careful algebra tells usa2+b2=s2−2p,&amp;nbsp;a3+b2=s3−3sp,&amp;nbsp;and&amp;nbsp;a5+b5=s5−5s3p+5sp2If&amp;nbsp;3(a5+b5)=11(a3+b3), then3s5−11s3−(15s3−33s)p+15sp2=0Now if&amp;nbsp;a2+b2=5, then&amp;nbsp;2p=s2−5, and the equation above (multiplied by 4) becomes12s5−44s3−2(15s3−33s)(s2−5)+15s(s2−5)2=0which simplifies to&amp;nbsp;0=3s5−22s3+45s=s(s2−9)(3s2+5)The real roots here are&amp;nbsp;s=0,3,and&amp;nbsp;−3&amp;nbsp;and these give values&amp;nbsp;p=−52, 2, and 2, respectively, for quadraticsx2−52,&amp;nbsp;x2−3x+2,&amp;nbsp;and&amp;nbsp;x2+3x+2The complex roots are&amp;nbsp;s=±53i, both of which give&amp;nbsp;p=−103, for quadraticsx2-53ix-103&amp;nbsp;&amp;nbsp;and&amp;nbsp;&amp;nbsp;x2+53ix-103

annieljcddj0 · Accepted Answer

If a+b=0, then you have a=±52,b=∓52.From this you can find the equation. If a+b≠0⇒3(a4−a3b+a2b2−ab3+b4). Using =11(a2−ab+b2)=11(5−ab)a4+b4=(a2+b2)2−2a2b2=25−2(ab)2,a3b+ab3=ab(a2+b2)=5abThus you do have a quadratic equation in ab, and once you solve for ab, then you can solve for a,b and you are done !

Find the quadratic equation from given relatioship The quadratic

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