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Question: If $\alpha, \beta, \gamma$ are the roots of the equation $2x^3 - 5x^2 + 4x - 3 = 0$, then $\sum \alp...

If α,β,γ\alpha, \beta, \gamma are the roots of the equation 2x3−5x2+4x−3=02x^3 - 5x^2 + 4x - 3 = 0, then ∑αβ(α+β)\sum \alpha \beta (\alpha + \beta)

Answer

1/2

Explanation

Solution

Given the cubic equation

2x3−5x2+4x−3=0,2x^3 - 5x^2 + 4x - 3 = 0,

with roots α,β,γ\alpha, \beta, \gamma, by Vieta’s formulas we have:

α+β+γ=52,αβ+βγ+γα=42=2,αβγ=32.\begin{aligned} &\alpha+\beta+\gamma = \frac{5}{2},\\[1mm] &\alpha\beta+\beta\gamma+\gamma\alpha = \frac{4}{2} = 2,\\[1mm] &\alpha\beta\gamma = \frac{3}{2}. \end{aligned}

We need to evaluate

∑αβ(α+β)=αβ(α+β)+βγ(β+γ)+γα(γ+α).\sum \alpha\beta(\alpha+\beta) = \alpha\beta(\alpha+\beta)+\beta\gamma(\beta+\gamma)+\gamma\alpha(\gamma+\alpha).

Notice that:

αβ(α+β)=α2β+αβ2,\alpha\beta(\alpha+\beta) = \alpha^2\beta + \alpha\beta^2,

so the sum becomes:

α2β+αβ2+β2γ+βγ2+γ2α+γα2.\alpha^2\beta + \alpha\beta^2 + \beta^2\gamma + \beta\gamma^2 + \gamma^2\alpha + \gamma\alpha^2.

This symmetric sum can be directly expressed in terms of the elementary symmetric sums by the formula:

α2β+αβ2+β2γ+βγ2+γ2α+γα2=(α+β+γ)(αβ+βγ+γα)−3αβγ.\alpha^2\beta + \alpha\beta^2 + \beta^2\gamma + \beta\gamma^2 + \gamma^2\alpha + \gamma\alpha^2 = (\alpha+\beta+\gamma)(\alpha\beta+\beta\gamma+\gamma\alpha) - 3\alpha\beta\gamma.

Substitute the values:

∑αβ(α+β)=(52)(2)−3(32)=5−92=10−92=12.\begin{aligned} \sum \alpha\beta(\alpha+\beta) &= \left(\frac{5}{2}\right)(2) - 3\left(\frac{3}{2}\right)\\[1mm] &= 5 - \frac{9}{2}\\[1mm] &= \frac{10-9}{2} = \frac{1}{2}. \end{aligned}