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:
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For polynomial inequalities, consider the discriminant condition carefully to determine valid intervals.
Step 1: Determining the condition for no real roots. The provided equation is: \[2x^2 + (a - 5)x + 15 = 3a\] Rearranging the equation yields: \[2x^2 + (a - 5)x + 15 - 3a = 0\] For the equation to have no real roots, the discriminant must be less than zero: \[(a - 5)^2 - 8(15 - 3a)<0\] Expanding the expression: \[a^2 + 25 - 10a - 120 + 24a<0\] Simplifying the inequality: \[a^2 + 14a - 95<0\] Factoring the quadratic: \[(a + 19)(a - 5)<0\] This inequality is satisfied for the following range of 'a': \[-19<a<5\]
Step 2: Identifying integer values within the determined interval. The integers 'a' such that -19<a<5 are: \[\{-18, -17, \ldots, -1, 0, 1, \ldots, 4\}\]
Step 3: Calculating the sum of the squares of these integer values. The sum of the squares is expressed as: \[\sum_{x \in X} x^2 = (1^2 + 2^2 + \cdots + 4^2) + (1^2 + 2^2 + \cdots + 18^2)\] Applying the formula for the sum of the first 'n' squares, $\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$: \[= \frac{4 \times 5 \times 9}{6} + \frac{18 \times 19 \times 37}{6}\] Evaluating the sums: \[= 30 + 2109 = 2139\]
