# Let, $\alpha$, $\beta$, $\gamma$ be the real roots of the equation, $x^7-3x^4+x=0$, where, $\alpha \gt \beta$. How is, $\alpha$ and $\beta$ related? [closed]

Let, $\alpha$, $\beta$, $\gamma$ be the 3 real roots of the equation, $x^7-3x^4+x=0$, where, $\alpha \gt \beta$, $\alpha \ne 0$, $\beta \ne 0$.

How is, $\alpha$ and $\beta$ related ?

Let, $\alpha=a$,$\beta=b$

The imaginary roots always occurs in pairs, so $\gamma$ must be equal to $0$.

$x^7-3x^4+x$

=$x(x-a)(x-b)(x^4+...)$

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## closed as too localized by BenjaLim, Davide Giraudo, Brandon Carter, rschwieb, Andres CaicedoJan 14 '13 at 16:19

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Do you know how to solve a quadratic equation? –  user38268 Jan 14 '13 at 14:10

Hint: having factored out one power of $x$, let $y=x^3$
To expand on Ross' hint, $$x^7-3x^4+x=x(x^6-3x^3+1)=x((x^3)^2-3(x^3)+1)$$ so you know that one root is $0=\gamma$. The other term is a quadratic in $x^3$ so by the quadratic formula $$x^3=\frac{3\pm\sqrt{5}}{2}$$ so $$\alpha=\left(\frac{3+\sqrt{5}}{2}\right)^{1/3}\qquad\beta=\left(\frac{3-\sqrt{5}}{2}\right)^{1/3}$$ Now come up with some relations between $\alpha$ and $\beta$. You might look among $\alpha^3, \beta^3$, and $\alpha\beta$, for example.