Question

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, $x\left(x + b^{3}\right) + \left(a^{3} ? 3abx\right)$ = 0 has roots :

• A. $\alpha^{3/2}$ and $\beta^{3/2}$
• B. $\alpha \,\beta^{1/2}$ and $\alpha^{1/2}\,\beta$
• C. $\sqrt{\alpha \,\beta }$ and $\alpha\,\beta$
• D. $\alpha^{-\frac{3}{2}}$ and $\beta^{-\frac{3}{2}}$

Answer 1

The Correct Option is A

To find the roots of the equation \( x(x + b^3) + (a^3 - 3abx) = 0 \), we start by examining the given quadratic equation \( ax^2 + bx + 1 = 0 \) where \(\frac{1}{\sqrt{\alpha}}\) and \(\frac{1}{\sqrt{\beta}}\) are its roots.

1. Using Vieta's formulas for the quadratic equation \( ax^2 + bx + 1 = 0 \), we know:
   • The sum of roots, \( \frac{1}{\sqrt{\alpha}} + \frac{1}{\sqrt{\beta}} = -\frac{b}{a} \)
   • The product of roots, \( \frac{1}{\sqrt{\alpha}} \cdot \frac{1}{\sqrt{\beta}} = \frac{1}{a} \)
2. From the product of roots, it follows that \(\frac{1}{\sqrt{\alpha \beta}} = \frac{1}{a}\) or \(\alpha \beta = a^2\).
3. The new equation \( x(x + b^3) + (a^3 - 3abx) = 0 \) simplifies to:
   • \( x^2 + b^3x + a^3 - 3abx = 0 \)
   • Re-organize to get \( x^2 + (b^3 - 3ab)x + a^3 = 0 \)

Let us find the roots of this new equation. Consider the transformations of the roots:

4. Suppose the roots of \( ax^2 + bx + 1 = 0 \) are \(\frac{1}{\sqrt{\alpha}}\) and \(\frac{1}{\sqrt{\beta}}\), implying that \(\frac{1}{x_i} = \sqrt{\alpha_{i}}\) for i = 1, 2.
5. When considering the roots of the transformed equation \((x_i^{1.5} = \alpha_{i})\), it results in roots of \( \alpha^{3/2} \) and \( \beta^{3/2} \).

Therefore, the correct option is \( \alpha^{3/2} \) and \( \beta^{3/2} \), which matches the given solution. This transformation aligns with the polynomial structure of the modified equation.