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 \( 1 \).

Answer 2

Given: \( a \) is a common root of the equations \[ x^2 - 5x + \lambda = 0, \] and \[ x^2 - 8x - 2\lambda = 0, \] with \( \lambda \neq 0 \). Let the other roots be \( \beta \) (from the first equation) and \( \gamma \) (from the second equation). Find the value of \[ a + \beta + \gamma + \lambda. \]

Step 1: Use Viète’s formulas for the first quadratic The roots of \[ x^2 - 5x + \lambda = 0 \] are \( a \) and \( \beta \). Sum of roots: \[ a + \beta = 5. \] Product of roots: \[ a \beta = \lambda. \]

Step 2: Use Viète’s formulas for the second quadratic The roots of \[ x^2 - 8x - 2\lambda = 0 \] are \( a \) and \( \gamma \). Sum of roots: \[ a + \gamma = 8. \] Product of roots: \[ a \gamma = -2 \lambda. \]

Step 3: Since \( a \) is a common root, it satisfies both equations Plug \( x = a \) into both equations: \[ a^2 - 5a + \lambda = 0 \quad \Rightarrow \quad \lambda = 5a - a^2, \] \[ a^2 - 8a - 2 \lambda = 0 \quad \Rightarrow \quad 2 \lambda = a^2 - 8a \quad \Rightarrow \quad \lambda = \frac{a^2 - 8a}{2}. \] Equate the two expressions for \( \lambda \): \[ 5a - a^2 = \frac{a^2 - 8a}{2}. \] Multiply both sides by 2: \[ 2(5a - a^2) = a^2 - 8a, \] \[ 10a - 2a^2 = a^2 - 8a, \] Bring all terms to one side: \[ 10a - 2a^2 - a^2 + 8a = 0, \] \[ 18a - 3a^2 = 0, \] \[ 3a^2 = 18a, \] \[ a^2 = 6a, \] \[ a^2 - 6a = 0, \] \[ a(a - 6) = 0. \] Since \( \lambda \neq 0 \), \( a \neq 0 \), so \[ a = 6. \]

Step 4: Find \( \lambda \) Using \( \lambda = 5a - a^2 \): \[ \lambda = 5 \times 6 - 6^2 = 30 - 36 = -6. \]

Step 5: Find \( \beta \) and \( \gamma \) From Step 1: \[ a + \beta = 5 \implies \beta = 5 - a = 5 - 6 = -1. \] From Step 2: \[ a + \gamma = 8 \implies \gamma = 8 - a = 8 - 6 = 2. \]

Step 6: Calculate \( a + \beta + \gamma + \lambda \) \[ 6 + (-1) + 2 + (-6) = 6 - 1 + 2 - 6 = 1. \]

Final answer: \[ \boxed{1}. \]