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Quadratic Equations

Section: Mathematics

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Question : 2 of 51

Marks: +1, -0

If

\alpha, \beta

\gamma

are the roots of the equation

2 x^3+3 x^2-5 x-7=0

\frac{1}{\alpha^2}+\frac{1}{\beta^2}+\frac{1}{\gamma^2}=

[AP EAPCET 24 May 2025 S1]

$-\frac{17}{49}$
$-\frac{23}{49}$
$\frac{55}{49}$
$\frac{67}{49}$

Solution:

Given, equation is

2 x^3+3 x^2-5 x-7=0

Since,

\alpha, \beta

and

\gamma

are the roots of the given equation.

\; \text{Therefore Sum of roots}\; =\alpha+\beta+\gamma=\frac{-3}{2}

Sum of products of roots

=\alpha\beta+\beta\gamma+\gamma\alpha=-\frac{5}{2}

Products of roots

=\alpha\beta\gamma=-\left(\frac{-7}{2}\right)=\frac{7}{2}

\frac{1}{\alpha^2}+\frac{1}{\beta^2}+\frac{1}{\gamma^2}=\frac{\beta^2\gamma^2+\gamma^2\alpha^2+\alpha^2\beta^2}{\alpha^2\beta^2\gamma^2}

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

= \frac{ (\alpha\beta+\beta\gamma+\gamma\alpha)^2 - 2\alpha\beta\gamma(\alpha+\beta+\gamma) }{ (\alpha\beta\gamma)^2 }

= \frac{ \left(\frac{-5}{2}\right)^2 - 2\left(\frac{7}{2}\right)\left(\frac{-3}{2}\right) }{ \left(\frac{7}{2}\right)^2 }

= \frac{ \frac{25}{4} + \frac{21}{2} }{ \frac{49}{4} } = \frac{ \frac{67}{4} }{ \frac{49}{4} }

= \frac{67}{4} \times \frac{4}{49} = \frac{67}{49}

\therefore \frac{1}{\alpha^2}+\frac{1}{\beta^2}+\frac{1}{\gamma^2}=\frac{67}{49}

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