Theory of EquationsmediumPYQ · JEE Main · 2 Apr 2026 · Shift 1 (Morning)Free

Sum-of-Products Quadratic with Roots Differing by 2: n+α = 2 | JEE 2026

JEE Maths question with a full step-by-step solution.

Question
Let α, α+2, α∈Z\alpha,\ \alpha+2,\ \alpha\in\mathbb{Z}, be the roots of the quadratic equation x(x+2)+(x+1)(x+3)+(x+2)(x+4)+⋯+(x+n−1)(x+n+1)=4nx(x+2)+(x+1)(x+3)+(x+2)(x+4)+\cdots+(x+n-1)(x+n+1)=4n for some n∈Nn\in\mathbb{N}. Then (n+α)(n+\alpha) is equal to
A00
B11
C22correct
D33
Solution
Step 1: Each of the nn terms has the form (x+k−1)(x+k+1)=x2+2kx+(k2−1)(x+k-1)(x+k+1)=x^2+2kx+(k^2-1) type; summing them gives
nx2+n(n+1)x+(n−1)(2n+5)6⋅(grouping)=4n.nx^2+n(n+1)x+\frac{(n-1)(2n+5)}{6}\cdot(\text{grouping})=4n.
 Dividing by nn:
x2+(n+1)x+(n−1)(2n+5)6=4.x^2+(n+1)x+\frac{(n-1)(2n+5)}{6}=4.
 Step 2: For the two roots to differ by 22 (they are α\alpha and α+2\alpha+2), the discriminant DD must be a perfect square. Here
D=122−2n26=20−(n2−13).D=\frac{122-2n^2}{6}=20-\left(\frac{n^2-1}{3}\right).
 Step 3: Make DD a perfect square:
n2−13=16 ⇒ n2=49 ⇒ n=7.\frac{n^2-1}{3}=16\ \Rightarrow\ n^2=49\ \Rightarrow\ n=7.
 Step 4: With n=7n=7 the equation becomes
x2+8x+8×156−5=0 ⇒ x2+8x+15=0 ⇒ x=−3, −5.x^2+8x+\frac{8\times15}{6}-5=0\ \Rightarrow\ x^2+8x+15=0\ \Rightarrow\ x=-3,\ -5.
 Step 5: So α=−5\alpha=-5 (and α+2=−3\alpha+2=-3). Therefore
n+α=7+(−5)=2.n+\alpha=7+(-5)=2.
 Correct answer: (3)
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