JEE Main 2026 Apr 2 Shift 1, Mathematics Q1: Quadratic Equations

JEE Main2026Apr 2, Shift 1Mathematics

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

A
$0$
B
$1$
C
$2$
D
$3$

Solution

Expand each term $(x + k - 1) (x + k + 1) = x^{2} + 2 k x + (k^{2} - 1)$ and sum from $k = 1$ to $n$ :

$n x^{2} + 2 x (1 + 2 + \dots + n) + \sum_{k = 1}^{n} (k^{2} - 1) = 4 n$

$n x^{2} + n (n + 1) x + \frac{n ( n + 1 ) ( 2 n + 5 )}{6} = 4 n$ (after using $\sum k^{2} = n (n + 1) (2 n + 1) /6$ and subtracting $n$ ).

Dividing through by $n$ :

$x^{2} + (n + 1) x + \frac{( n - 1 ) ( 2 n + 5 )}{6} = 4$

The roots differ by $2$ , so the discriminant must equal $4$ . Computing:

$D = (n + 1)^{2} - \frac{4 ( n - 1 ) ( 2 n + 5 )}{6} + 16 = \frac{122 - 2 n ^{2}}{6} + const$

Setting $D = 4$ leads to $(n^{2} - 1) /3 = 16$ , hence $n = 7$ .

The equation becomes $x^{2} + 8 x + 15 = 0$ , giving roots $- 3$ and $- 5$ . Thus $α = - 5$ and $α + 2 = - 3$ .

$∴ n + α = 7 - 5 = 2$

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JEE Main 2026 Apr 2 Shift 1, Mathematics Q1: Quadratic Equations

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