Question

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

• A. $a < \frac{1}{2}$
• B. $a >\frac{1}{2}$
• C. $a < 1$
• D. $a >\frac{11}{9}$

Answer 1

Answer: (D)

(i) $ D \geq 0$
$36 a^2-4\left(2-2 a+9 a^2\right) \geq 0 $
$36 a^2-8+8 a-36 a^2 \geq 0 $
$8 a \geq 8 \Rightarrow a \geq 1$
(ii)$f(3)>0 $
$9-18 a+2-2 a+9 a^2>0$
$9 a^2-20 a+11>0 $
$(a-1)(9 a-11)>0$
$a \in(-\infty,+1) \cup\left(\frac{11}{9}, 0\right)$

(iii) $\frac{-B}{2 A}>3 $
$\frac{6 a}{2}>3$
$a>1$
Intersection I, II and III $a \in\left(\frac{11}{9}, 0\right) \Rightarrow a >\frac{11}{9}$.