Question

Two roots of the equation $ax^4 + bx^3 + cx^2 + dx + e = 0$ are positive and equal. If the product of the other two real roots is 1, then

• A. $be^2 = a^2d$
• B. $\frac{3e + 2b\sqrt{e} + c}{\sqrt{a}} = a$
• C. $e + 2b\sqrt{e} + 3c = a\sqrt{a}$
• D. $b^2e = ad^2$

Hint:

For polynomials with repeated roots and specific conditions (e.g., $\beta\gamma = 1$), use Vieta’s formulas and substitute to derive relationships between coefficients.

Answer 1

The Correct Option is D

Let the roots of $ax^4 + bx^3 + cx^2 + dx + e = 0$ be $\alpha, \alpha, \beta, \gamma$, with $\alpha>0$ and $\beta\gamma = 1$. 
For a quartic $ax^4 + bx^3 + cx^2 + dx + e = 0$, Vieta’s formulas give: 
- Sum of roots: $\alpha + \alpha + \beta + \gamma = 2\alpha + \beta + \gamma = -\frac{b}{a}$ - Product of roots: $\alpha \cdot \alpha \cdot \beta \cdot \gamma = \alpha^2 \cdot 1 = \alpha^2 = \frac{e}{a}$ 
- Sum of pairwise products: $\alpha^2 + \alpha\beta + \alpha\gamma + \alpha\beta + \alpha\gamma + \beta\gamma = \alpha^2 + 2\alpha(\beta + \gamma) + 1 = \frac{c}{a}$ 
- Sum of triple products: $\alpha^2(\beta + \gamma) + \alpha(\beta\gamma) + \alpha(\beta\gamma) = \alpha^2(\beta + \gamma) + 2\alpha = -\frac{d}{a}$ From $\beta\gamma = 1$, let $\gamma = \frac{1}{\beta}$. Then: \[ 2\alpha + \beta + \frac{1}{\beta} = -\frac{b}{a} \quad (1) \] \[ \alpha^2 = \frac{e}{a} \implies e = a\alpha^2 \quad (2) \] \[ \alpha^2 + 2\alpha\left(\beta + \frac{1}{\beta}\right) + 1 = \frac{c}{a} \quad (3) \] \[ \alpha^2\left(\beta + \frac{1}{\beta}\right) + 2\alpha = -\frac{d}{a} \quad (4) \] From (4), let $s = \beta + \frac{1}{\beta}$: \[ \alpha^2 s + 2\alpha = -\frac{d}{a} \implies d = -a(\alpha^2 s + 2\alpha) \quad (5) \] Square (5): \[ d^2 = a^2 (\alpha^2 s + 2\alpha)^2 = a^2 \alpha^2 (s^2 \alpha^2 + 4s\alpha + 4) \] From (1): $s = -\frac{b}{a} - 2\alpha$. Compute: \[ b^2 = a^2 (2\alpha + s)^2 = a^2 (4\alpha^2 + 4\alpha s + s^2) \] \[ b^2 e = a^2 (4\alpha^2 + 4\alpha s + s^2) \cdot a \alpha^2 = a^3 \alpha^2 (4\alpha^2 + 4\alpha s + s^2) \] Compare with $a d^2$: \[ a d^2 = a \cdot a^2 \alpha^2 (s^2 \alpha^2 + 4s\alpha + 4) = a^3 \alpha^2 (s^2 \alpha^2 + 4s\alpha + 4) \] Since $s^2 \alpha^2 + 4s\alpha + 4 = (s\alpha + 2)^2$ and $4\alpha^2 + 4\alpha s + s^2 = (2\alpha + s)^2$, and both are equal from (1), we have: \[ b^2 e = a d^2 \] Option (4) is correct. Options (1), (2), and (3) do not hold under the given conditions.