Find the cubic equation whose roots are twice those of the equation 3x^3 -2x^2 +1=0



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aruv's profile pic

Posted on (Answer #1)


`alpha,beta and gamma `  are roots of the equation`3x^3-2x^2+1=0`



`alpha beta+beta gamma +gamma alpha=0`

`alpha beta gamma=(-1)/3`

Since required equation has roots `2alpha ,2 beta and 2gamma`


`2 alpha+2 beta+2 gamma=2(alpha+beta+gamma)=2xx2/3=4/3`

`2alpha 2beta+2 beta 2 gamma+2 gamma 2 alpha=`

`4(alpha beta +beta gamma+gama alpha)=4xx0=0`

`2alpha 2 beta 2 gamma=8 alpha beta gamma=-8/3`

Thus required equation is



degeneratecircle's profile pic

Posted on (Answer #2)

Let `r` be a solution of `3x^3-2x^2+1=0,` so that `3r^3-2r^2+1=0.`

In that case, `2r` is a root of `3(x/2)^3-2(x/2)^2+1=0,` which simplifies to `3/8x^3-x^2/2+1=0.`

There is no "the" cubic equation with such roots--in other words, it's not unique. Any nonzero multiple of the last equation will suffice, for example if we multiply it by 8 we get


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