Find all values of $a$ for which both roots of the equation ${ x }^{ 2 }-6ax+2-2a+9{ a }^{ 2 }=0$ are greater than $3$.
- For all $a\in \left( { 11 }/{ 9 },+\infty \right) $
- For all $a\in \left( { 22 }/{ 9 },+\infty \right) $
- For all $a\in \left( { 11 }/{ 3 },+\infty \right) $
- For all $a\in \left( { 1 }/{ 9 },+\infty \right) $
For both roots > 3: 1) Discriminant >= 0: 36a^2 - 4(9a^2-2a+2) >= 0 -> 8a - 8 >= 0 -> a >= 1. 2) Vertex > 3: -b/2a = 6a/2 = 3a > 3 -> a > 1. 3) f(3) > 0: 9 - 18a + 9a^2 - 2a + 2 > 0 -> 9a^2 - 20a + 11 > 0 -> (9a-11)(a-1) > 0. For a > 1, this holds if a > 11/9.
To find the values of a where both roots are greater than 3, we must satisfy three conditions: the discriminant must be non-negative, the vertex (axis of symmetry) must be greater than 3, and the function evaluated at x = 3 must be positive. The discriminant is (6a)^2 - 4(9a^2 - 2a + 2) = 8a - 8, which requires a >= 1. The axis of symmetry is x = 6a/2 = 3a, which requires a > 1. Finally, evaluating the polynomial at x = 3 gives 9 - 18a + 2 - 2a + 9a^2 > 0, simplifying to 9a^2 - 20a + 11 > 0, which factors into (9a - 11)(a - 1) > 0. The intersection of these conditions requires a to be strictly greater than 11/9. The result is for all a in (11/9, +infinity).