Question

If \( \alpha \neq \beta \), \( \alpha^2 = 5\alpha - 3 \), \( \beta^2 = 5\beta - 3 \), then the equation having \( \frac{\alpha}{\beta} \) and \( \frac{\beta}{\alpha} \) as its roots is

• A. \( 3x^2 - 19x - 3 = 0 \)
• B. \( 3x^2 + 19x - 3 = 0 \)
• C. \( x^2 + 19x + 3 = 0 \)
• D. \( 3x^2 - 19x - 19 = 0 \)
• E. \( 3x^2 - 19x + 3 = 0 \)

Hint:

Use identities \( \alpha^2+\beta^2 = (\alpha+\beta)^2 - 2\alpha\beta \) to simplify expressions.

Answer 1

Answer: (E)

Concept: Given: \[ \alpha^2 - 5\alpha + 3 = 0 \] Thus roots are \( \alpha, \beta \). \[ \alpha + \beta = 5, \quad \alpha\beta = 3 \]

Step 1: Find sum of new roots.
\[ \frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha\beta} \] \[ \alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta = 25 - 6 = 19 \] \[ \Rightarrow = \frac{19}{3} \]

Step 2: Find product.
\[ \frac{\alpha}{\beta} \cdot \frac{\beta}{\alpha} = 1 \]

Step 3: Form equation.
\[ x^2 - \frac{19}{3}x + 1 = 0 \] Multiply by 3: \[ 3x^2 - 19x + 3 = 0 \]