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4-2.Quadratic Equations and Inequations
normal
$\alpha$, $\beta$ ,$\gamma$ are roots of equatiuon $x^3 -x -1 = 0$ then equation whose roots are $\frac{1}{{\beta + \gamma }},\frac{1}{{\gamma + \alpha }},\frac{1}{{\alpha + \beta }}$ is
A
$x^3 -x^2 + 1 = 0$
B
$x^3 + x^2 -1 = 0$
C
$x^3 + x -1 = 0$
D
$x^3 -x + 1 = 0$
Solution
$x+\beta+\gamma=0$
$\therefore \alpha+\beta=-\gamma, \beta+\gamma=-\alpha, \gamma+\alpha=-\beta$
$\therefore$ Roots of Reqd. equation are $\frac{-1}{\alpha}, \frac{-1}{\beta}, \frac{-1}{\gamma}$
-ve recipical roots,
$\therefore \text { Replace } \mathrm{x} \rightarrow \frac{-1}{\mathrm{x}}$
$\frac{-1}{x^{3}}+\frac{1}{x}-1=0 \quad \Rightarrow-1+x^{2}-x^{3}=0$
$\text { or } x^{3}-x^{2}+1=0$
Standard 11
Mathematics
