If both the roots of a quadratic equation $x ^ { 2 } - 6 a x + \left(2 - 2 a + 9 a ^ { 2 }\right) = 0$ exceed $3$ then
- $a > \dfrac { 9 } { 11 }$
- $a \geq \dfrac { 11 } { 9 }$
- $a > \dfrac { 11 } { 9 }$
- $a < \dfrac { 11 } { 9 }$
For roots of x^2 - 6ax + (2-2a+9a^2) = 0 to exceed 3: 1) Discriminant >= 0: 36a^2 - 4(2-2a+9a^2) >= 0 => 36a^2 - 8 + 8a - 36a^2 >= 0 => 8a >= 8 => a >= 1. 2) Vertex > 3: 6a/2 > 3 => 3a > 3 => a > 1. 3) f(3) > 0: 9 - 18a + 2 - 2a + 9a^2 > 0 => 9a^2 - 20a + 11 > 0. Roots of 9a^2 - 20a + 11 are 1 and 11/9. Inequality holds for a < 1 or a > 11/9. Combining conditions, a > 11/9.
Let f(x) = x^2 - 6ax + 2 - 2a + 9a^2. For both roots to exceed 3, three conditions must be met: the sum of the roots must exceed 6, f(3) must be positive, and the vertex must be past 3. The sum condition gives 6a > 6, so a > 1. The vertex condition gives -(-6a)/2 > 3, which also means a > 1. Evaluating f(3) > 0 gives 9 - 18a + 2 - 2a + 9a^2 > 0, simplifying to 9a^2 - 20a + 11 > 0, which factors to (9a - 11)(a - 1) > 0. Combining these conditions, the intersection gives a > 11/9.