Question

If both the roots of the equation $x^2-6ax+2-2a+9a^2=0exceed3,$then

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

Answer 1

The correct option is D

$a>\frac{11}{9}$

Explantion for the correct options:

Finding the value:

Let $\alpha and\beta$be the roots of the given equation , then $\alpha +\beta =6aand\alpha \beta =9a^2-2a+2$ given that the roots exceed by $3.$

The roots become $\alpha +3and\beta +3$.

$$\begin{gathered} \mathit{A}\mathit{M}\mathbf{\ge }\mathit{G}\mathit{M} \\ \Rightarrow \frac{\left[\alpha +3+\beta +3\right]}{2}\ge \sqrt{(\alpha +3)(\beta +3)} \\ \Rightarrow \frac{\left[\alpha +\beta +6\right]}{2}\ge \sqrt{\alpha \beta +3\alpha +3\beta +9} \\ \Rightarrow \frac{\left[6a+6\right]}{2}\ge \sqrt{9a^2-2a+2+3\left(6a\right)+9} \\ \Rightarrow 3a+3\ge \sqrt{9a^2+6a+11} \\ \Rightarrow 9{\left(a+1\right)}^2\ge 9a^2+16a+11 \\ \Rightarrow 9a^2+18a+9\ge 9a^2+16a+11 \\ \Rightarrow \mathbf{2}\mathit{a}\mathbf{\ge }\mathbf{2} \\ \Rightarrow \mathit{a}\mathbf{\ge }\mathbf{1} \end{gathered}$$

Hence option(D) is correct.