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If \alpha ,\beta ,\gamma are roots of x\left ( 1+x^{2} \right )+x^{2}\left ( 6+x \right )+2=0  then \alpha ^{-1}+\beta ^{-1}+\gamma ^{-1}  equals

  • Option 1)

    -1

  • Option 2)

    \frac{-1}{2}

  • Option 3)

    0

  • Option 4)

    \frac{1}{2}

 

Answers (1)

best_answer

Equation becomes :  2x^{3}+6x^{2}+x+2=0

\because \alpha ,\beta ,\gamma ,\; are\; roots\; ,\; so

\alpha \beta +\beta \gamma +\gamma \alpha =\frac{1}{2}\; and\; \alpha \beta \gamma =-1\\*\therefore \frac{1}{\alpha }+\frac{1}{\beta }+\frac{1}{\gamma }=\frac{\beta \gamma +\alpha \gamma +\alpha \beta }{\alpha \beta \gamma }=\frac{\frac{1}{2}}{-1}=\frac{-1}{2}

 

Sum of product of pair of roots in cubic equation -

\alpha \beta +\beta \gamma +\gamma \alpha = \frac{c}{a}

- wherein

ax^{3}+bx^{2}+cx+d= 0

is the cubic equation

 

 


Option 1)

-1

This is incorrect

Option 2)

\frac{-1}{2}

This is correct

Option 3)

0

This is incorrect

Option 4)

\frac{1}{2}

This is incorrect

Posted by

prateek

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