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 $a\in R$, for which the equation $2{x}^{2}+(a-5)x+15=3a$ has no real root, is the interval ( $\alpha ,\beta $ ), and $X=|x\in Z;\alpha <x<\beta |$, then $\sum _{x\in X}{x}^{2}$ is equal to:

Accepted Answer

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

Solutions · 1