Forming New Polynomials using properties of roots

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Paradoxica
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Forming New Polynomials using properties of roots

Postby Paradoxica » Tue Mar 10, 2015 9:23 am UTC

x3 + px - 1 = 0 has three real, non-zero roots (α,β,γ)
a) Find the values of α2 + β2 + γ2 and α4 + β4 + γ4 in terms of p, and hence show that p is negative.
b) Find the monic cubic equation with co-efficients in terms of p, whose roots are α2, β2, γ2
I've completed part a), and almost completed part b), but my problem is that the solution is x3 + 2px2 - 1 = 0
So far, I have managed to construct a cubic with those roots, but I have x3 + 2px2 + p2x - 1 = 0.
I'm not sure if the answer is wrong or not (as per usual)
a) α+β+γ = -b/a = 0
α2 + β2 + γ2 = (α+β+γ)2 - 2(αβ+βγ+γα) = -2p
α2 + β2 + γ2 > 0
Sum of squares of real non-zeroes must be strictly positive
-2p > 0
p < 0
Not sure how sum of fourth powers helps, since this can be deduced from sum of squares.
b) let y = x2 and hence, we will deduce the new polynomial from substitution
x3 + px - 1 = 0
x3 + px = 1
x(x2 + p) = 1
x2(x2 + p)2 = 1
y(y + p)2 = 1
y3 + 2py2 + p2y = 1
y3 + 2py2 + p2y -1 = 0
replacing y with x for the required polynomial with roots α2, β2, γ2
x3 + 2px2 + p2x - 1 = 0
Last edited by Paradoxica on Tue Mar 10, 2015 9:12 pm UTC, edited 1 time in total.
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Qaanol
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Re: Forming New Polynomials using properties of roots

Postby Qaanol » Tue Mar 10, 2015 5:57 pm UTC

Please show the work you’ve done so far.
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Qaanol
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Re: Forming New Polynomials using properties of roots

Postby Qaanol » Tue Mar 10, 2015 9:45 pm UTC

Thank you for showing your work. If you had made a new post rather than editing your existing one then I would have seen right away that you had done so.

Now, how could you verify which of those polynomials has the correct roots?
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Paradoxica
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Re: Forming New Polynomials using properties of roots

Postby Paradoxica » Wed Mar 11, 2015 2:01 am UTC

Strongly suspect the answer is wrong. Used Sum and Product to get the same polynomial I got earlier, and still couldn't figure out how to eliminate p2x from the equation.
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Re: Forming New Polynomials using properties of roots

Postby phlip » Wed Mar 11, 2015 2:31 am UTC

What happens if you just try it? Pick a value for p, find the roots (approximately is good enough), build the polynomial with those roots squared, and see what it matches?

It's by no means a proof that either answer is correct, but it'll work to show that one of the answers is wrong...

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notzeb
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Re: Forming New Polynomials using properties of roots

Postby notzeb » Wed Mar 11, 2015 2:39 am UTC

phlip wrote:What happens if you just try it? Pick a value for p, find the roots (approximately is good enough), build the polynomial with those roots squared, and see what it matches?

It's by no means a proof that either answer is correct, but it'll work to show that one of the answers is wrong...
Fun side problem: it turns out there is no value of p for which all three roots are rational numbers. (If you take p = -2 you get pretty nice roots, even if they aren't rational.)

Edit: Also, Descartes' rule of signs tells us that x3 + 2px2 - 1 can't possibly be the answer: it doesn't have enough positive roots.
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