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

### 2 Answers | Add Yours

Let

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

Thus

`alpha+beta+gamma=2/3`

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

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

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

Thus

`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

`x^3-(4/3)x^2+0x-(-8/3)=0`

`3x^3-4x^2+8=0`

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

`3x^3-4x^2+8=0.`

### Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes