Mathematics · Quadratic Equations

JEE Main 2025 — 29 January, Evening Shift — Question 42

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 $\sum_{x \in X} x^2 \text{ is }$

Option A:
2109
Option B:
2126
Option C:
2139
Correct
Option D:
2119

Answer: C

Step-by-step solution

(a-5)^2-8(15-3a)<0

a^2-10a+25-120+24a<0

a^2+14a-95<0

(a+19)(a-5)<0

$\Rightarrow -19 <\ a <\ 5$

Let $X$ be the set of integers satisfying the inequality. Then,

X=\{-18,-17,\ldots,-1,0,1,2,3,4\}

\sum_{x\in X}x^2 = (1^2+2^2+\cdots+18^2)+(1^2+2^2+\cdots+4^2)

Using

\sum_{k=1}^{n}k^2=\frac{n(n+1)(2n+1)}{6}

\sum_{k=1}^{4}k^2=\frac{4\cdot5\cdot9}{6}=30

\sum_{k=1}^{18}k^2=\frac{18\cdot19\cdot37}{6}=2109

\sum_{x\in X}x^2=30+2109={2139}

{\sum_{x\in X}x^2={2139}}

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 29 January, Evening Shift
Subject: Mathematics
Chapter: Quadratic Equations
Topic: Nature of Roots of Quadratic Equation