I'm having trouble with this problem.
Show that if the roots of $$5x^3-x^2-2x+3=0$$ are $a_1,a_2,a_3$, then $$1/a_1+1/a_2+1/a_3=2/3$$
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2$1/a_1+1/a_2+1/a_3=(a_2 a_3 +a_1 a_3+a_1 a_2)/(a_1 a_2 a_3)$ and http://en.wikipedia.org/wiki/Vieta%27s_formulas – Cortizol Nov 21 '14 at 15:27
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1The Fundamental Theorem of Algebra says something different. – Dietrich Burde Nov 21 '14 at 15:32
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I'm supposed to use FTA to solve it. I forgot to mention that. – user183782 Nov 21 '14 at 15:39
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1The FTA has nothing to do with the solution of this problem. Viète's formulas are completely independent of it; if you assume the existence of the roots, you don't need to find them, do you? – egreg Nov 21 '14 at 17:19
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The equation $3y^3-2y^2-y+5=0$ has roots the reciprocals of the roots of $5x^3-x^2-2x+3=0$. To see this, divide the given polynomial through by $x^3$, and reverse the order of summation.
The sum of the roots of $ 3y^3-2y^2-y+5=0$ is $-\frac{-2}{3}$.
beep-boop
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André Nicolas
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Use that since $a_1,a_2,a_3$ are the roots of the polynomial then $$5x^3-x^2-2x+3=5(x-a_1)(x-a_2)(x-a_3)$$ Expand the right side and compare the coefficients of equal powers of $x$ to obtain that $$\begin{cases}a_1a_2+a_1a_3+a_2a_3=-\dfrac{2}{5}\\a_1a_2a_3=-\dfrac{3}{5}\end{cases}$$ Now $$\frac{a_1a_2+a_1a_3+a_2a_3}{a_1a_2a_3}=\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}$$ gives you the result.
Jimmy R.
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@user183782 Again with coefficient comparison you know that $(a_1+a_2+a_3)=1$ and therefore $(a_1+a_2+a_3)^2=1/25$ which implies that $$a_1^2+a_2^2+a_3^2=1/25-2(a_1a_2+a_1a_3+a_2a_3)=1/25-2(-2/5)=1/25+20/25=21/25$$ as you want. – Jimmy R. Nov 21 '14 at 16:57
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$$\frac{1}{a_1}+\frac{1}{a_2} + \frac{1}{a_2} = \frac{a_2a_3+a_1a_3+a_1a_2}{a_1a_2a_3} = \frac{(-2)/5}{(-1)^3 \cdot 3/5}=\frac{2}{3}$$
Mher
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1You might want to explain how you got values for a1, a2, and a3...rather than a link – stackErr Nov 21 '14 at 15:36