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

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Aicharya · Accepted Answer

$\begin{aligned} & (a-5)^2-8(15-3 a) < 0 \ & a^2+14 a+25-120 < 0 \ & a^2+14 a-95 < 0 \ & (a+19)(a-5) < 0 \ & a \in(-19,5) \ & 	herefore-19 < x < 5 \ & 	herefore \sum_{x \in X} x^2=\left(1^2+2^2+\ldots .+4^2ight)+\left(1^2+2^2+\ldots+18^2ight) \ & =\frac{4 	imes 5 	imes 9}{6}+\frac{18 	imes 19 	imes 37}{6} \ & =30+2109 \ & =2139\end{aligned}$ ~

If the set of all $\mathrm{a} \in \mathbf{R}$, for which the equation $2 x^2+(a-5) x+15=3 \mathrm{a}$ has no…

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