Let α ,α +2,α in Z be the roots of the quadratic equation x ( x + 2) + ( x + 1)( x + 3) + ( x + 2)( x + 4) + ldots +( x + n - 1)( x + n + 1) = 4 n for some n in N . Then ( n + α) is equal to:

Mathematics · Quadratic Equations

JEE Main 2026 — 2 April, Morning Shift — Question 20

Let $\alpha ,\alpha +2,\alpha \in \mathbf{Z}$ be the roots of the quadratic equation $\mathbf{x}(\mathbf{x} + 2) + (\mathbf{x} + 1)(\mathbf{x} + 3) + (\mathbf{x} + 2)(\mathbf{x} + 4) + \ldots +(\mathbf{x} + \mathbf{n} - 1)(\mathbf{x} + \mathbf{n} + 1) = 4\mathbf{n}$ for some $\mathbf{n}\in \mathbf{N}$ . Then $(\mathbf{n} + \alpha)$ is equal to:

Option A:
0
Option B:
1
Option C:
2
Correct
Option D:
3

Answer: C

Step-by-step solution

$\mathrm{nx}^{2}+\mathrm{x}(2+4+6+\ldots . .+2 \mathrm{n})+(1.3+\ldots . .+(\mathrm{n}-1) (\mathrm{n}+1))=4 \mathrm{n}$ $n x^{2}+n(n+1) x+\frac{n(n-1)(2 n+5)}{6}=4 n$ $\mathrm{x}^{2}+(\mathrm{n}+1) \mathrm{x}+\frac{(\mathrm{n}-1)(2 \mathrm{n}+5)}{6}=4$ D must be a perfect square $\mathrm{D}=\frac{122-2 \mathrm{n}^{2}}{6}=20-\left(\frac{\mathrm{n}^{2}-1}{3}\right)$ If make perfect square $\Rightarrow \frac{\mathrm{n}^{2}-1}{3}=16 \Rightarrow \mathrm{n}=7$ So Equation is $\Rightarrow \mathrm{x}^{2}+8 \mathrm{x}+\frac{8 \times 15}{6}-5=0$ $\mathrm{x}^{2}+8 \mathrm{x}+15=0$ $\mathrm{x}=-3,-5$ $\alpha=-5, \alpha+2=-3$ $\alpha+\mathrm{n}=7-5=2$

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Exam: JEE Main 2026
Paper: 2 April, Morning Shift
Subject: Mathematics
Chapter: Quadratic Equations
Topic: Theory of Quadratic Equations