How would I prove that the sum of the squares of the roots of $x^3+ax^2+bx+c$ is equal to $a^2-2b$?
How can I prove sum of the squares of the roots of $x^3+ax^2+bx+c$ is equal to $a^2-2b$?
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algebra-precalculus
polynomials
roots
cubic-equations
symmetric-polynomials
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0Do you know how to express the coefficients in terms of the roots. If not you can look it up. It should help. – 2017-02-24
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1https://en.wikipedia.org/wiki/Vieta's_formulas should be helpful. – 2017-02-24
1 Answers
6
Say the roots are $r,s,t$. Then $$(x-s)(x-r)(x-t)=x^3+ax^2+bx+c$$
Comparing coefficients we see that $$s+r+t=-a\quad \& \quad sr+st+rt=b$$
It follows that $$a^2=(s+r+t)^2=s^2+r^2+t^2+2(sr+st+rt)$$ and we are done.