If both the roots of the equation x^2 - 6ax + | vedclass

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(D) Given equation: $x^2 - 6ax + (9a^2 - 2a + 2) = 0$. Let $f(x) = x^2 - 6ax + 9a^2 - 2a + 2$. For both roots to be greater than $3$,the following con

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

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(D) Given equation: $x^2 - 6ax + (9a^2 - 2a + 2) = 0$. Let $f(x) = x^2 - 6ax + 9a^2 - 2a + 2$. For both roots to be greater than $3$,the following conditions must be satisfied: $1)$ Discriminant $D \ge 0$: $D = (-6a)^2 - 4(1)(9a^2 - 2a + 2) = 36a^2 - 36a^2 + 8a - 8 = 8a - 8$. $8a - 8 \ge 0 \Rightarrow a \ge 1$. $2)$ Vertex location: $-\frac{b}{2a} > 3$: $-\frac{-6a}{2(1)} > 3$ $\Rightarrow 3a > 3$ $\Rightarrow a > 1$. $3)$ $f(3) > 0$: $f(3) = 3^2 - 6a(3) + 9a^2 - 2a + 2 > 0$ $9 - 18a + 9a^2 - 2a + 2 > 0$ $9a^2 - 20a + 11 > 0$ $(9a - 11)(a - 1) > 0$. Since $a > 1$,the condition $(9a - 11)(a - 1) > 0$ implies $a > \frac{11}{9}$. Combining all conditions,we get $a > \frac{11}{9}$.

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

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