- Home
- Standard 11
- Mathematics
4-2.Quadratic Equations and Inequations
hard
If the set of all $a \in R$, for which the equation $2 x^2+$ $(a-5) x+15=3 a$ 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
A$2109$
B$2129$
C$2139$
D$2119$
(JEE MAIN-2025)
Solution
$(a-5)^2-8(15-3 a)<0$
$a^2+14 a+25-120<0$
$a^2+14 a-95<0$
$(a+19)(a-5)<0$
$a \in(-19,5)$
$\therefore-19$
$\therefore \sum_{x \in X } x^2=\left(1^2+2^2+\ldots .+4^2\right)+\left(1^2+2^2+\ldots+18^2\right)$
$=\frac{4 \times 5 \times 9}{6}+\frac{18 \times 19 \times 37}{6}$
$=30+2109$
$=2139$
$a^2+14 a+25-120<0$
$a^2+14 a-95<0$
$(a+19)(a-5)<0$
$a \in(-19,5)$
$\therefore-19$
$\therefore \sum_{x \in X } x^2=\left(1^2+2^2+\ldots .+4^2\right)+\left(1^2+2^2+\ldots+18^2\right)$
$=\frac{4 \times 5 \times 9}{6}+\frac{18 \times 19 \times 37}{6}$
$=30+2109$
$=2139$
Standard 11
Mathematics
