Question

If the set of all \( a \in \mathbb{R} \), for which the equation \( 2x^2 + (a - 5)x + 15 = 3a \) has no real root, is the interval \( (\alpha, \beta) \), and \( X = \{ x \in \mathbb{Z} : \alpha<x<\beta \} \), then \( \sum_{x \in X} x^2 \) is equal to:

• A. 2109
• B. 2129
• C. 2139
• D. 2119

Hint:

For polynomial inequalities, consider the discriminant condition carefully to determine valid intervals.

Answer 1

The Correct Option is C

Step 1: Identifying the condition for no real roots. The given equation is: \[ 2x^2 + (a - 5)x + 15 = 3a \] Rearranging, \[ 2x^2 + (a - 5)x + 15 - 3a = 0 \] For no real roots, the discriminant must be negative: \[ (a - 5)^2 - 8(15 - 3a)<0 \] Expanding, \[ a^2 + 25 - 10a - 120 + 24a<0 \] \[ a^2 + 14a - 95<0 \] \[ (a + 19)(a - 5)<0 \] This inequality holds true for: \[ -19<a<5 \]

Step 2: Finding integers in this interval. The integer values between -19 and 5 are: \[ \{-18, -17, \ldots, -1, 0, 1, \ldots, 4\} \]

Step 3: Summing the squares of the values. \[ \sum_{x \in X} x^2 = (1^2 + 2^2 + \cdots + 4^2) + (1^2 + 2^2 + \cdots + 18^2) \] Using the sum of squares formula: \[ \sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6} \] \[ = \frac{4 \times 5 \times 9}{6} + \frac{18 \times 19 \times 37}{6} \] \[ = 30 + 2109 = 2139 \]