JEE Main 2026 — Quadratic Equation Question with Solution

The given quadratic equation is $(n^2-2n+2)x^2-3x+(n^2-2n+2{)}^2=0$.

Let $k=n^2-2n+2=(n-1{)}^2+1$. Since $(n-1{)}^2\ge 0$ for all $n\in \mathbb{R}$, the minimum value of $k$ is $1$ at $n=1$.

The product of the roots is given by $P=\frac{(n^2-2n+2{)}^2}{n^2-2n+2}=n^2-2n+2=k$.

The minimum value of the product of the roots is $\alpha =\min (k)=1$.

The sum of the roots is given by $S=\frac{3}{n^2-2n+2}=\frac{3}{k}$.

The maximum value of the sum of the roots occurs when $k$ is minimum. Thus, $\beta =\max \left(\frac{3}{k}\right)=\frac{3}{1}=3$.

We are given a Geometric Progression (G.P.) whose first term is $a=\alpha =1$ and common ratio is $r=\frac{\alpha }{\beta }=\frac{1}{3}$.

The sum of the first six terms of this G.P. is:

$S_6=a\frac{1-r^6}{1-r}=1\cdot \frac{1-{\left(\frac{1}{3}\right)}^6}{1-\frac{1}{3}}$

$S_6=\frac{1-\frac{1}{729}}{\frac{2}{3}}=\frac{\frac{728}{729}}{\frac{2}{3}}$

$S_6=\frac{728}{729}\times \frac{3}{2}=\frac{364}{243}$

Answer: $\frac{364}{243}$