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4-2.Quadratic Equations and Inequations
medium
If $\alpha ,\,\beta ,\,\gamma $ are the roots of the equation ${x^3} + 4x + 1 = 0,$ then ${(\alpha + \beta )^{ - 1}} + {(\beta + \gamma )^{ - 1}} + {(\gamma + \alpha )^{ - 1}} = $
A
$2$
B
$3$
C
$4$
D
$5$
Solution
(c) If $\alpha, \beta , \gamma $ are the roots of the equation.
${x^3} – p{x^2} + qx – r = 0$
$\therefore {(\alpha + \beta )^{ – 1}} + {(\beta + \gamma )^{ – 1}} + {(\gamma + \alpha )^{ – 1}} = \frac{{{p^2} + q}}{{pq – r}}$
Given, $p = 0,\,\,q = 4,\,\,\,r = – 1$
==> $\frac{{{p^2} + q}}{{pq – r}} = \frac{{0 + 4}}{{0 + 1}} = 4$.
Standard 11
Mathematics
