Question

Find all values of $a$ for which both roots of the equation $x^2-6ax+2-2a+9a^2=0$ are greater than $3$.

• A. For all $a\in (11/9,+\infty )$
• B. For all $a\in (22/9,+\infty )$
• C. For all $a\in (11/3,+\infty )$
• D. For all $a\in (1/9,+\infty )$

Answer 1

The correct option is A For all $a\in (11/9,+\infty )$

$x^2-6ax+2-2a+9a^2=0$
Condition for both roots to be greater than 3, following three conditions must be satisfied
(i) $D\ge 0$
$\Rightarrow 36a^2-36a^2+8a+8\ge 0$
$\Rightarrow a\ge -1$
(iii) $-\frac{B}{2A}<3$
$\Rightarrow \frac{6a}{1}<3$
$\Rightarrow a<3$
(iii) $Af\left(3\right)>0$
$9-18a+2-2a+9a^2>0$
$\Rightarrow 9a^2-20a+11>0$
$\Rightarrow a\in (-\infty ,1)\cup (\frac{11}{9},\infty )$
From (i),(ii) and (iii), we get
$a\in (\frac{11}{9},\infty )$