JEE Mains · Maths · STD 11 - 4.2 Quadratic equations and inequations
If \(\frac{1}{{\sqrt \alpha }}\) and \(\frac{1}{{\sqrt \beta }}\) are the roots of the equation , \(ax^2 + bx + 1 = 0\) \(\left( {a \ne 0,a,b \in R} \right)\), then the equation, \(x ( x + b^3 ) + (a^3 - 3abx ) = 0\) has roots
- A \({\alpha ^{\frac{3}{2}}}\) and \({\beta ^{^{\frac{3}{2}}}}\)
- B \(\alpha {\beta ^{\frac{1}{2}}}\) and \({\alpha ^{^{\frac{1}{2}}}}\beta \)
- C \(\sqrt {\alpha \beta } \) and \(\alpha \beta \)
- D \({\alpha ^{ - \frac{3}{2}}}\) and \({\beta ^{^{ - \frac{3}{2}}}}\)
Answer & Solution
Correct Answer
(A) \({\alpha ^{\frac{3}{2}}}\) and \({\beta ^{^{\frac{3}{2}}}}\)
Step-by-step Solution
Detailed explanation
Let \(\frac{1}{{\sqrt \alpha }}\) and \(\frac{1}{{\sqrt \beta }}\) be the roots of \(ax^{2}+bx+1=0\) \(\frac{1}{\sqrt{\alpha}}+\frac{1}{\sqrt{\beta}}\) \(=\left(\frac{\sqrt{\alpha}+\sqrt{\beta}}{\sqrt{\alpha \beta}}\right)\) \(=-\frac{b}{a}\)…
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