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Study Stuff for MCA Examinations - NIMCET
Study Stuff for MCA Examinations - NIMCET
Previous Year Question (PYQs)
If $\alpha$ and $\beta$ are the two roots of the quadratic equation $x^2 + ax + b = 0, (ab \ne 0)$ then the quadratic roots whose roots $\frac{1}{\alpha^3+\alpha}$ and $\frac{1}{\beta^3+\beta}$ is
Solution
Given $x^{2} + ax + b = 0$ with roots $\alpha,\beta$,we use
$\alpha + \beta = -a$ and $\alpha\beta = b$.
Required new roots are
$\dfrac{1}{\alpha^{3}+\alpha}$ and $\dfrac{1}{\beta^{3}+\beta}$.
Since
$\alpha^{3}+\alpha = \alpha(\alpha^{2}+1)$
and using $\alpha^{2}=-a\alpha-b$ (and same for $\beta$),
after simplification the sum and product of new roots become:
$u+v = \dfrac{a^{3}+a-3ab}{b(b^{2}+1+a^{2}-2b)}$
$uv = \dfrac{1}{b(b^{2}+1+a^{2}-2b)}$
So the required quadratic is:
$b(b^{2}+1+a^{2}-2b)x^{2} - (a^{3}+a-3ab)x + 1 = 0$
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