Question

If \( a \) is a common root of \( x^2 - 5x + \lambda = 0 \) and \( x^2 - 8x - 2\lambda = 0 \) (\( \lambda \neq 0 \)) and \( \beta, \gamma \) are the other roots of them, then \( a + \beta + \gamma + \lambda = \):

• A. 0
• B. -1
• C. 1
• D. 2

Hint:

For problems involving common roots of quadratic equations, use Vieta’s formulas to relate the sum and product of the roots, and solve the resulting system of equations.

Answer 1

The Correct Option is C

We are given that \( a \) is a common root of the equations: \[ x^2 - 5x + \lambda = 0 \quad \text{and} \quad x^2 - 8x - 2\lambda = 0. \] Using Vieta's formulas for both quadratic equations, we can write the following relations for the sum and product of the roots: 1. For \( x^2 - 5x + \lambda = 0 \): - Sum of roots: \( a + \beta = 5 \), - Product of roots: \( a \beta = \lambda \). 2. For \( x^2 - 8x - 2\lambda = 0 \): - Sum of roots: \( a + \gamma = 8 \), - Product of roots: \( a \gamma = -2\lambda \). Now, solving the system of equations: - From the sum of roots in the first equation, we get \( \beta = 5 - a \). - From the sum of roots in the second equation, we get \( \gamma = 8 - a \). Now substitute these expressions for \( \beta \) and \( \gamma \) into the product relations: - \( a(5 - a) = \lambda \), - \( a(8 - a) = -2\lambda \). Now, solve these equations: 1. \( a(5 - a) = \lambda \) gives: \[ 5a - a^2 = \lambda. \] 2. \( a(8 - a) = -2\lambda \) gives: \[ 8a - a^2 = -2\lambda. \] Substitute \( \lambda = 5a - a^2 \) into the second equation: \[ 8a - a^2 = -2(5a - a^2), \] \[ 8a - a^2 = -10a + 2a^2, \] \[ 8a - a^2 + 10a - 2a^2 = 0, \] \[ 18a - 3a^2 = 0, \] \[ a(18 - 3a) = 0. \] Thus, \( a = 0 \) or \( a = 6 \). Step 1: Determine \( a + \beta + \gamma + \lambda \) If \( a = 6 \), then from the previous equations: \[ \beta = 5 - 6 = -1, \quad \gamma = 8 - 6 = 2. \] Thus, \( a + \beta + \gamma = 6 - 1 + 2 = 7 \). And using \( \lambda = 5a - a^2 = 5(6) - 6^2 = 30 - 36 = -6 \), we get: \[ a + \beta + \gamma + \lambda = 7 - 6 = 1. \] Thus, the value of \( a + \beta + \gamma + \lambda \) is \( 1 \).