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

If both the roots of the equation $x^2-6 a x+2-2 a+9 a^2=0$ exceeds 3 , then
  1. $a \lt \frac{3}{2}$
  2. $a\gt\frac{3}{2}$
  3. $a \lt \frac{5}{2}$
  4. $a\gt\frac{11}{9}$

Solution

$\begin{aligned} & x^2-6 a x+2-2 a+9 a^2=0 \\ & \because \quad \mathrm{D}\gt0 \Rightarrow 36 a^2-4\left(2-2 a+9 a^2\right)\gt0 \\ & \Rightarrow 8(a-1)\gt0 \Rightarrow a\gt1 \end{aligned}$ $f(3)\gt0 \Rightarrow 9-18 a-2 a+9 a^2+2\gt0$ $\begin{aligned} & \Rightarrow 9 a^2-20 a+11\gt0 \\ & \Rightarrow(9 a-11)(a-1)\gt0 \Rightarrow a\gt\frac{11}{9} \\ & f^{\prime}(3) \lt 0 \Rightarrow 6-6 a \lt 0 \Rightarrow a\gt1 \end{aligned}$
Condition for roots to be greater than 3 is $a\gt\frac{11}{9}$.

Asked in: AP EAMCET 2024 (22 May Shift 1)

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