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TS EAMCET 19-July-2022 Shift 2 Solved Paper

Section: Mathematics

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Question : 12 of 160

Marks: +1, -0

If

\alpha, \beta, \gamma

are the roots of the equation

5 x^3-3 x^2+2 x-4=0

\sum \alpha^2 \beta^2=

$\;\frac{4}{5}$
$\;\frac{-2}{5}$
$\;\frac{2}{5}$
$\;\frac{-4}{5}$

Solution:

Given,

\alpha, \beta

and

\gamma

are the roots of the equation

5 x^3-3 x^2+2 x-4=0

Then,

\alpha+\beta+\gamma=\;\frac{3}{5},\ \alpha \beta+\beta \gamma+\gamma \alpha=\;\frac{2}{5}

and

\alpha \beta \gamma=\;\frac{4}{5}

Now,

\alpha+\beta+\gamma=\;\frac{3}{5}

On squaring both sides, we get

\;(\alpha+\beta+\gamma)^2=\;\frac{9}{25}

\Rightarrow \alpha^2+\beta^2+\gamma^2+2(\alpha \beta+\beta \alpha+\gamma \alpha)=\;\frac{9}{25}

\Rightarrow \alpha^2+\beta^2+\gamma^2=\;\frac{9}{25}-\;\frac{4}{5}=-\;\frac{11}{25}

\therefore \;\alpha \beta+\beta \gamma+\gamma \alpha=\;\frac{2}{5}

On squaring both sides, we get

\;\alpha^2 \beta^2+\beta^2 \gamma^2+\alpha^2 \gamma^2+2\alpha \beta^2 \gamma+2\alpha \beta \gamma^2+2\alpha^2 \beta \gamma=\;\frac{4}{25}

\;\Rightarrow \sum \alpha^2 \beta^2+2 \alpha \beta \gamma(\alpha+\beta+\gamma)=\frac{4}{25}

\;\Rightarrow \sum \alpha^2 \beta^2=\;\frac{4}{25}-2\left(\;\frac{4}{5}\right)\left(\;\frac{3}{5}\right)=\;\frac{4}{25}-\;\frac{24}{25}=-\;\frac{20}{25}=-\;\frac{4}{5}

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