Mathematics · 2014
JEE Main · 9 April 2014, Shift 1 · Q64
If 1/(√ α) and 1/(√ β) are the roots of the equation, a x^2+ b x+1=0( a ≠ 0, a, b ∈ R ), then the equation, x(x+ b^3)+( a^3-3 ab x)=0 has roots:
If $\displaystyle \frac{1}{\sqrt{\alpha}}$ and $\displaystyle \frac{1}{\sqrt{\beta}}$ are the roots of the equation, $\displaystyle \mathrm{a} x^2+\mathrm{b} x+1=0(\mathrm{a} \neq 0, \mathrm{a}, \mathrm{b} \in \mathbf{R})$, then the equation, $\displaystyle x\left(x+\mathrm{b}^3\right)+\left(\mathrm{a}^3-3 \mathrm{ab} x\right)=0$ has roots :
Official answer
(1)
$\displaystyle \alpha^{3 / 2}$ and $\displaystyle \beta^{3 / 2}$
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JEE Main 2014 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.