Consider the quadratic equation (n^2-2n+2)x^2-3x+(n^2-2n+2)^2=0, nR.

Question

Consider the quadratic equation $(n^2-2n+2)x^2-3x+(n^2-2n+2)^2=0$, $n\in\mathbb{R}$. Let $\alpha$ be the minimum value of the product of its roots and $\beta$ be the maximum value of the sum of its roots. Then the sum of the first six terms of the G.P., whose first term is $\alpha$ and common ratio is $\dfrac{\alpha}{\beta}$, is:

Accepted Answer

Let $a = n^2-2n+2 = (n-1)^2+1 \ge 1$. With $b=-3$, $c=a^2$:
$$\text{Product of roots} = \frac{c}{a} = a, \qquad \text{Sum of roots} = \frac{3}{a}.$$

Since $a\ge1$ (equality at $n=1$), the minimum product is $\alpha=1$, and the maximum sum occurs at the same minimum $a=1$, giving $\beta = 3$.

G.P.: first term $\alpha=1$, ratio $r=\alpha/\beta = 1/3$.

$$S_6 = \frac{1-(1/3)^6}{1-1/3} = \frac{728/729}{2/3} = \frac{364}{243}.$$

Solutions · 1