Aspire Faculty ID #11973 · Topic: NIMCET 2025 · Just now
NIMCET 2025

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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