Algebra – Quadratic Equations – JEE Main 21 Jan 2026 Shift 2

Solution:

Given equation can be written using summation:
$$\sum_{r=1}^{n}(x+2r-2)(x+2r)=\frac{8n}{3}$$

$$\sum_{r=1}^{n}(x^2 + x(2r-2+2r) + (2r-2)2r)=\frac{8n}{3}$$

$$\sum_{r=1}^{n}(x^2 + 2x(2r-1) + 4r(r-1))=\frac{8n}{3}$$

Using standard summation formulas $\sum_{r=1}^n (2r-1) = n^2$ and $\sum_{r=1}^n r(r-1) = \frac{n(n^2-1)}{3}$:
$$nx^{2}+2x(n^2)+4\left(\frac{n(n^2-1)}{3}\right)=\frac{8n}{3}$$

Divide the entire equation by $n$ ($n \neq 0$):
$$x^{2}+2nx+\frac{4(n^{2}-1)}{3}-\frac{8}{3}=0$$

$$x^{2}+2nx+\frac{4n^{2}-12}{3}=0$$

Let roots be $\alpha, \beta$. Since they are consecutive even integers, $|\alpha – \beta| = 2$.
Using difference of roots formula $|\alpha – \beta| = \frac{\sqrt{D}}{|a|}$:
$$\frac{\sqrt{(2n)^2 – 4(1)(\frac{4n^2-12}{3})}}{1} = 2$$

Squaring both sides:
$$4n^2 – \frac{16n^2-48}{3} = 4$$

$$12n^2 – (16n^2-48) = 12$$

$$-4n^2 + 48 = 12$$

$$4n^2 = 36 \Rightarrow n^2 = 9$$

$$n = 3 \quad (\text{Since } n \text{ is a positive integer})$$

Ans. (1)