Question

If \( \alpha, \beta, \gamma, \delta \) are the roots of the equation \( x^4 - 10x^3 + 35x^2 - 50x + 24 = 0 \) and \( \alpha^2, \beta^2, \gamma^2, \delta^2 \) are the roots of the equation \( x^4 + ax^3 + bx^2 + cx + d = 0 \), then \( a = \)

• A. \( 36 \)
• B. \( 30 \)
• C. \( -30 \)
• D. \( -36 \)

Hint:

Use the identity \( \sum \alpha_i^2 = (\sum \alpha_i)^2 - 2\sum \alpha_i\alpha_j \). Here, \( a = -(10^2 - 2(35)) = -(100 - 70) = -30 \). This works even if the roots aren't easy to find.

Answer 1

The Correct Option is C

Concept: The coefficient of the second highest power in a monic polynomial is the negative sum of its roots.
• For the first equation: \( \sum \alpha = 10 \), \( \sum \alpha\beta = 35 \).
• For the second equation: \( a = -(\alpha^2 + \beta^2 + \gamma^2 + \delta^2) \).

Step 1: Identifying the roots of the first equation.
Notice that \( x^4 - 10x^3 + 35x^2 - 50x + 24 = (x-1)(x-2)(x-3)(x-4) \). The roots are \( 1, 2, 3, 4 \).

Step 2: Calculating the sum of the squares of the roots.
The roots of the second equation are \( 1^2, 2^2, 3^2, 4^2 \), which are \( 1, 4, 9, 16 \). Sum of new roots \( = 1 + 4 + 9 + 16 = 30 \).

Step 3: Finding \( a \).
By Vieta's formulas, \( a = -(\text{Sum of new roots}) = -30 \).