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Question 1199426: let the roots of the equation x^3 -2x^2 -3x-7=0 be r, s, and t. find the equation whose roots are r^2, s^2 and t^2
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
r, s, and t are roots of a cubic.
Which means x-r, x-s, x-t are factors.
Multiply those factors and expand like so.
(x-r)(x-s)(x-t)
(x-r)(x^2-tx-sx+st)
x(x^2-tx-sx+st)-r(x^2-tx-sx+st)
(x^3-tx^2-sx^2+stx)+(-rx^2+rtx+rsx-rst)
x^3-tx^2-sx^2+stx-rx^2+rtx+rsx-rst
x^3+(-tx^2-sx^2-rx^2)+(stx+rtx+rsx)-rst
x^3-(t+s+r)x^2+(st+rt+rs)x-rst
x^3-(r+s+t)x^2+(rs+st+rt)x-rst
In short
(x-r)(x-s)(x-t) = x^3-(r+s+t)x^2+(rs+st+rt)x-rst
This is mentioned in Vieta's formulas regarding cubics.
Compare
x^3-(r+s+t)x^2+(rs+st+rt)x-rst
with
x^3-2x^2-3x-7
to see that- r+s+t = 2
- rs+st+rt = -3
- rst = 7
Through similar calculations, we'll have:
(x-r^2)(x-s^2)(x-t^2) = x^3-(r^2+s^2+t^2)x^2+(r^2s^2+s^2t^2+r^2t^2)x-(rst)^2
We need to know the following three items:- r^2+s^2+t^2
- r^2s^2+s^2t^2+r^2t^2
- rst
to be able to determine the cubic equation that has roots r^2, s^2, and t^2.
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One useful polynomial identity is
(r+s+t)^2 = r^2+s^2+t^2+2(rs+st+rt)
The proof of which I leave to the reader.
That identity rearranges to
r^2+s^2+t^2 = (r+s+t)^2 - 2(rs+st+rt)
Then we plug in the items mentioned earlier
r^2+s^2+t^2 = (r+s+t)^2 - 2(rs+st+rt)
r^2+s^2+t^2 = (2)^2 - 2(-3)
r^2+s^2+t^2 = 10
With similar steps we'll have
(rs+st+rt)^2 = r^2s^2+s^2t^2+r^2t^2+2(rs^2t+rst^2+r^2st)
r^2s^2+s^2t^2+r^2t^2 = (rs+st+rt)^2 - 2(rs^2t+rst^2+r^2st)
r^2s^2+s^2t^2+r^2t^2 = (rs+st+rt)^2 - 2rst(r+s+t)
r^2s^2+s^2t^2+r^2t^2 = (-3)^2 - 2*7*(2)
r^2s^2+s^2t^2+r^2t^2 = -19
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We found the following- r^2+s^2+t^2 = 10
- r^2s^2+s^2t^2+r^2t^2 = -19
- rst = 7
and now have enough information.
Let's plug in those items to get the following.
x^3-(r^2+s^2+t^2)x^2+(r^2s^2+s^2t^2+r^2t^2)x-(rst)^2
x^3-10x^2+(-19)x-(7)^2
x^3-10x^2-19x-49
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Answer:
If x^3-2x^2-3x-7=0 has roots {r,s,t}, then the equation that has roots {r^2,s^2,t^2} is x^3-10x^2-19x-49=0
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