If the set of all a R, for which the equation 2x^2+(a-5)x+15=3a has no real root, is the i…

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 $\displaystyle\sum_{x\in X} x^2$ is equal to:

Accepted Answer

Rewrite as $2x^2+(a-5)x+(15-3a)=0$. No real root means $D<0$:
$$(a-5)^2 - 8(15-3a) < 0 \implies a^2-10a+25-120+24a<0 \implies a^2+14a-95<0$$
$$(a+19)(a-5)<0 \implies a \in (-19,5)$$

So $\alpha=-19,\ \beta=5$, and $X=\{x\in\mathbb Z: -19<x<5\} = \{-18,-17,\dots,4\}$.

$$\sum_{x\in X}x^2 = \left(\sum_{x=1}^{18}x^2\right) + \left(\sum_{x=1}^{4}x^2\right) = \frac{18\cdot19\cdot37}{6} + \frac{4\cdot5\cdot9}{6} = 2109+30 = 2139$$

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