Quadratic Equations – Nature of Roots – JEE Main 29 Jan 2025 Shift 2

Solution:

For the equation $2x^2 + (a-5)x + (15-3a) = 0$ to have no real roots, the discriminant must be less than zero ($D < 0$).
$(a-5)^2 – 4(2)(15-3a) < 0$
$a^2 – 10a + 25 – 8(15-3a) < 0$
$a^2 – 10a + 25 – 120 + 24a < 0$
$a^2 + 14a – 95 < 0$
$(a+19)(a-5) < 0$
$a \in (-19, 5)$

Thus, $\alpha = -19$ and $\beta = 5$.

The set $X = \{x \in Z : -19 < x < 5\} = \{-18, -17, \dots, 0, \dots, 4\}$.
The required sum is $\sum_{x \in X} x^2 = ((-18)^2 + \dots + (-1)^2) + 0^2 + (1^2 + \dots + 4^2)$

$= \sum_{n=1}^{18} n^2 + \sum_{n=1}^{4} n^2$

Using $\sum n^2 = \frac{n(n+1)(2n+1)}{6}$:

$= \frac{18 \times 19 \times 37}{6} + \frac{4 \times 5 \times 9}{6}$

$= 3(19)(37) + 30$

$= 2109 + 30 = 2139$

Ans. (3)