Solving Problems Using Vieta’s Formula

Posted: September 9, 2024 in Mathematics
Tags: math, Math Contest, Mathematics, MathOlympiad

Problem:

Solution:

By Vieta’s formula,

$\alpha+\beta+\gamma=1, \quad \alpha\beta+\beta\gamma+\gamma\alpha=0, \quad \alpha\beta\gamma=2$

Consider the following,

$(\alpha+\beta+\gamma)^2 = \alpha^2+\beta^2+\gamma^2+2(\alpha\beta+\beta\gamma+\gamma\alpha)$

hence,

$\alpha^2+\beta^2+\gamma^2=1$

Now, consider the following two identities,

$(\alpha^2+\beta^2+\gamma^2)^2 = \alpha^4+\beta^4+\gamma^4+2(\alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2)$

$(\alpha\beta+\beta\gamma+\gamma\alpha)^2=\alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2+2\alpha\beta\gamma(\alpha+\beta+\gamma)$

and hence the following,

$\alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2 = 0-2(2)(1) = -4$

$\alpha^4+\beta^4+\gamma^4 = 1 -2(-4)=9$

Now,

$\alpha^5+\beta^5+\gamma^5 = \alpha^3\alpha^2 + \beta^3\beta^2 + \gamma^3\gamma^2$

$= (\alpha^2+2)\alpha^2 + (\beta^2+2)\beta^2 + (\gamma^2+2)\gamma^2$

$= \alpha^4+\beta^4+\gamma^4+2(\alpha^2+\beta^2+\gamma^2)$

$= 9 + 2(1) = 11$

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