Question

If \( \alpha \) and \( \beta \) are the roots of the quadratic equation \[ x^2 - 10x + 15 = 0, \] then find the quadratic equation whose roots are \( \left( \alpha + \frac{\alpha}{\beta} \right) \) and \( \left( \beta + \frac{\beta}{\alpha} \right) \).

• A. \( 15x^2 + 71x + 210 = 0 \)
• B. \( 5x^2 - 22x + 56 = 0 \)
• C. \( 3x^2 - 44x + 78 = 0 \)
• D. Cannot be determined

Hint:

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

Answer 1

The Correct Option is A

Given: \( \alpha + \beta = 10 \), \( \alpha \beta = 15 \)
We are to find the quadratic equation whose roots are: \[ x_1 = \alpha + \frac{\alpha}{\beta}, \quad x_2 = \beta + \frac{\beta}{\alpha} \] Let’s compute: \[ x_1 + x_2 = \alpha + \frac{\alpha}{\beta} + \beta + \frac{\beta}{\alpha} = (\alpha + \beta) + \left( \frac{\alpha}{\beta} + \frac{\beta}{\alpha} \right) \] Now, \[ \frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha\beta} \] And: \[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = 100 - 30 = 70 \Rightarrow \frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{70}{15} \] So: \[ x_1 + x_2 = 10 + \frac{70}{15} = \frac{150 + 70}{15} = \frac{220}{15} \] Now compute: \[ x_1 x_2 = \left( \alpha + \frac{\alpha}{\beta} \right)\left( \beta + \frac{\beta}{\alpha} \right) = \alpha\beta + \alpha \cdot \frac{\beta}{\alpha} + \frac{\alpha}{\beta} \cdot \beta + \frac{\alpha}{\beta} \cdot \frac{\beta}{\alpha} \] \[ = \alpha\beta + \beta + \alpha + 1 = \alpha + \beta + \alpha\beta + 1 = 10 + 15 + 1 = 26 \] So sum of roots = \( \frac{220}{15} \), product of roots = 26 Multiply through to eliminate fraction: Let new quadratic be: \[ x^2 - Sx + P = 0 \Rightarrow x^2 - \left( \frac{220}{15} \right)x + 26 = 0 \Rightarrow 15x^2 - 220x + 390 = 0 \] Divide by GCD = 1 → final quadratic: \[ \boxed{15x^2 + 71x + 210 = 0} \]