Question

If both the roots of the quadratic equation $x^2-6nx+(9n^2-2n+2)=0$ are greater than 3, then the range of $n$ is:

• A. $(\frac{11}{9},\infty )$
• B. $(-\infty ,\frac{11}{9})$
• C. $[1,\infty )$
• D. $(-\infty ,1)$

Answer 1

The correct option is A $(\frac{11}{9},\infty )$

Given: $x^2-6nx+(9n^2-2n+2)=0$ Roots are greater than 3
To find: Range of $n$
Step$1:$ Draw the graph for the given expression $y=x^2-6nx+(9n^2-2n+2)$
Step$2:$ Write the applicable conditions.
Step$3:$ Solve all the conditions.
Step$4:$ Take intersection and combine the results for $n$.


Now, the applicable conditions that needs to be satisfied are:
$i).D\ge 0$
$ii).f(k)>0$
$iii).-\frac{b}{2a}>k$

Now, solving for all the conditions,
$i).D\ge 0$
$D=b^2-4ac\ge 0\Rightarrow (-6n{)}^2-4\cdot 1\cdot (9n^2-2n+2)\ge 0\Rightarrow 36n^2-36n^2+8n-8\ge 0\Rightarrow 8n-8\ge 0\Rightarrow 8n\ge 8\Rightarrow n\ge 1\Rightarrow n\in [1,\infty )\cdots (A)$

$ii).f(k)>0,$ where k is 3
$\Rightarrow f\left(3\right)>0$
$\Rightarrow 3^2-6n\times 3+(9n^2-2n+2)>0$
$\Rightarrow 9-18n+9n^2-2n+2>0$
$\Rightarrow 9n^2-20n+11>0$
$\Rightarrow (9n-11)(n-1)>0$
$\Rightarrow n<1\text{or}n>\frac{11}{9}$
$\Rightarrow n\in (-\infty ,1)\text{or}n\in (\frac{11}{9},\infty )$
$\Rightarrow n\in (-\infty ,1)\cup (\frac{11}{9},\infty )\cdots \left(B\right)$

$iii.)-\frac{b}{2a}>k$
$\Rightarrow \frac{6n}{2\cdot 1}>3$
$\Rightarrow 3n>3\Rightarrow n>1$
$\Rightarrow n\in (1,\infty )\cdots \left(C\right)$
$\therefore$ Range of $n$ will be the intersection of solution sets of $\left(A\right),\left(B\right)\&\left(C\right)$
$\Rightarrow n\in [1,\infty )\cap \{(-\infty ,1)\cup (\frac{11}{9},\infty )\}\cap (1,\infty )$
$\Rightarrow n\in (\frac{11}{9},\infty )$