StemCET Logo

Maths Question 1 – JEE-MAIN 2026

← Back
Let α,α+2,α∈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)=4n for some n∈N. Then n+α is equal to :

Recognize the pattern in the terms of the given equation and express the sum using summation notation.

Step 1: Simplify the summation into a quadratic equation✦ Active

The given equation is a sum of terms (x+k)(x+k+2) for k from 0 to n−1. The general term expands to x2+(2k+2)x+k2+2k. Summing these terms:

∑k=0n−1(x2+(2k+2)x+k2+2k)=nx2+x∑k=0n−1(2k+2)+∑k=0n−1(k2+2k)

Using summation formulas ∑k=0mk=m(m+1)2 and ∑k=0mk2=m(m+1)(2m+1)6 (with m=n−1):

∑k=0n−1(2k+2)=2(n−1)n2+2n=n(n+1) ∑k=0n−1(k2+2k)=(n−1)n(2n−1)6+2(n−1)n2=n(n−1)(2n+5)6

Substituting these back into the equation and dividing by n (since n∈N):

nx2+n(n+1)x+n(n−1)(2n+5)6=4n x2+(n+1)x+((n−1)(2n+5)6−4)=0
Step 2: Apply Vieta's formulas to find n○ Expand

Let the roots of the quadratic equation be α and α+2. Using Vieta's formulas:

Sum of roots: α+(α+2)=−(n+1)⟹2α+2=−n−1⟹2α=−n−3⟹α=−n−32. Since α∈Z, n must be an odd integer.

Product of roots: α(α+2)=(n−1)(2n+5)6−4. Substitute the expression for α:

(−n−32)(−n−32+2)=(n−1)(2n+5)6−4 (−n−32)(−n+12)=(n−1)(2n+5)6−4 (n+3)(n−1)4=(n−1)(2n+5)6−4

Multiply by 12 to clear denominators and simplify:

3(n+3)(n−1)=2(n−1)(2n+5)−48 3(n2+2n−3)=2(2n2+3n−5)−48 3n2+6n−9=4n2+6n−10−48 0=n2−49 n2=49⟹n=±7

Since n∈N, we take n=7. This is consistent with n being odd.

💡 Teacher's Secret Hint

Remember to check the domain constraints for n and α (e.g., n∈N, α∈Z). This helps validate the solution.

Step 3: Calculate n+α○ Expand

With n=7, we find α using the relation from the sum of roots:

α=−n−32=−7−32=−102=−5

Finally, calculate n+α:

n+α=7+(−5)=2
✦ STEM Console utilizes AI models to generate step-by-step explanations and math clues. AI can make mistakes.