Let the set of all $a \in \mathbb{R}$ for which the equation $2x^2 + (a-5)x + 15 = 3a$ has no real root be $(\alpha,\beta)$. If $X = \{x \in \mathbb{Z} : \alpha < x < \beta\}$, then $\displaystyle\sum_{x \in X} x^2$ is
Step-by-Step Solution
Key Concept: Set the discriminant of $2x^2+(a-5)x+(15-3a)=0$ strictly negative; solve the resulting quadratic inequality in $a$ to find $\alpha$ and $\beta$, then sum $x^2$ over all integers in the open interval.
Rewrite: $2x^2+(a-5)x+(15-3a)=0$. For no real root, discriminant $< 0$:
$$(a-5)^2 - 8(15-3a) < 0$$
$$a^2 - 10a + 25 - 120 + 24a < 0$$
$$a^2 + 14a - 95 < 0$$
$$(a+19)(a-5) < 0 \Rightarrow a \in (-19,\,5).$$
So $\alpha=-19$, $\beta=5$, and $X = \{-18,-17,\ldots,-1,0,1,2,3,4\}$.
$$\sum_{x\in X}x^2 = \sum_{k=1}^{18}k^2 + \sum_{k=1}^{4}k^2 = \frac{18\cdot19\cdot37}{6}+\frac{4\cdot5\cdot9}{6} = 2109+30 = 2139.$$
Correct Answer: 4