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

Question

Consider the quadratic equation $({n}^{2}-2n+2){x}^{2}-3x+{({n}^{2}-2n+2)}^{2}=0,n\in R$. 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 the common ratio is $\frac{\alpha }{\beta }$, is :

Accepted Answer

For the quadratic equation

$({n}^{2}-2n+2){x}^{2}-3x+{({n}^{2}-2n+2)}^{2}=0,$

let us first compare it with the standard quadratic form

$a{x}^{2}+bx+c=0.$

So,

$a={n}^{2}-2n+2,\, b=-3,\, c={({n}^{2}-2n+2)}^{2}.$

Now rewrite $a$ in a simpler form:

${n}^{2}-2n+2=(n-1{)}^{2}+1.$

Hence,

$a=(n-1{)}^{2}+1.$

Since $(n-1{)}^{2}\ge 0$, we get

$a\ge 1.$

Now, for a quadratic equation, by Vieta's formulas:

Sum of roots $=-\frac{b}{a}$

Product of roots $=\frac{c}{a}$

So here,

$=\frac{3}{a}$

and

$=\frac{{a}^{2}}{a}=a.$

Therefore,

$=(n-1{)}^{2}+1.$

So the minimum value of the product is clearly

$\alpha =1,$

attained when $n=1$.

Now the sum of roots is

$\frac{3}{(n-1{)}^{2}+1}.$

Since the denominator is minimum when $(n-1{)}^{2}=0$, the maximum value of the sum is

$\beta =\frac{3}{1}=3.$

So we have

$\alpha =1,\, \beta =3.$

Now the G.P. has

first term $a=\alpha =1$

common ratio $r=\frac{\alpha }{\beta }=\frac{1}{3}$

We need the sum of first six terms:

${S}_{6}=a\cdot \frac{1-{r}^{6}}{1-r}.$

Substitute the values:

${S}_{6}=1\cdot \frac{1-{(\frac{1}{3})}^{6}}{1-\frac{1}{3}}.$

Now,

${(\frac{1}{3})}^{6}=\frac{1}{729}$

and

$1-\frac{1}{3}=\frac{2}{3}.$

Hence,

${S}_{6}=\frac{1-\frac{1}{729}}{\frac{2}{3}}=\frac{\frac{728}{729}}{\frac{2}{3}}=\frac{728}{729}\cdot \frac{3}{2}.$

So,

${S}_{6}=\frac{1092}{729}=\frac{364}{243}.$

Therefore, the required sum is

$\frac{364}{243}.$

So the correct option is

$.$

Solutions · 1