Question

If $\alpha \neq \beta$, but $\alpha^{2} = 5\alpha - 3,\beta^{2} = 5\beta - 3$, then the equation whose roots are $\frac{\alpha}{\beta}$ and $\frac{\beta}{\alpha}$ is

• A. $x^{2} - 5x - 3 = 0$
• B. $3x^{2} - 19x + 3 = 0$
• C. $3x^{2} + 12x + 3 = 0$
• D. None of these

Answer 1

Answer: (B)

$3x^{2} - 19x + 3 = 0$

Answer 2

$S = \frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^{2} + \beta^{2}}{\alpha\beta} = \frac{5\alpha - 3 + 5\beta - 3}{\alpha\beta}$ $\left\lbrack \begin{aligned} & \because\alpha^{2} = 5\alpha - 3 \\ & \beta^{2} = 5\beta - 3 \end{aligned} \right\rbrack$

$S = \frac{5(\alpha + \beta) - 6}{\alpha\beta}$, $p = \frac{\alpha}{\beta} \cdot \frac{\beta}{\alpha} = 1$ ⇒ $p = 1$. α, β are roots of $x^{2} - 5x + 3 = 0$. Therefore $\alpha + \beta = 5$, $\alpha\beta = 3$

$S = \frac{5(5) - 6}{3} = \frac{19}{3}$

∵ $x^{2} - \frac{19}{3}x + 1 = 0$ ⇒ $3x^{2} - 19x + 3 = 0$