Question

If both the roots of the equation $x^2-6ax+2-2a+9a^2=0$ exceed 3, then find the value of a.

• A. undefined
• B. undefined
• C. undefined
• D. undefined

Answer 1

$x^2-6ax+2-2a+9a^2=0$, for both the roots to be greater than 3,
Graph is concave upward since $a>0$ (+ve)

For both roots to exceed 3,
(i) $f\left(3\right)>0$
(ii) $D\ge 0$
(i) $f\left(3\right)>0\Rightarrow (3{)}^2-6a(3)+2a-2a+9a^2>0\Rightarrow 9a^2-20a+11>0$
$\Rightarrow 9a^2-9a-11a+11>0\Rightarrow 9a(a-1)-11(a-1)>0$ $\Rightarrow (9a-11)(a-1)>0$
$\Rightarrow a>1anda>\frac{11}{9};aϵ(-\infty ,1)\cup (\frac{11}{9},\infty )$

(ii) $D\ge 0\Rightarrow 36a^2-4(2-2a+9a^2)\ge 0\Rightarrow -8+8a\ge 0\Rightarrow a\ge 1$
Combining (i) and (ii); $a>\frac{11}{9}\to$ Condition on a for roots to be greater than 3.