For what values of $m\in R$, both roots of the equation $x^2-6mx+9m^2-2m+2=0$ exceed $3$?
- $[-7, -2]$
- $\left(\dfrac {11}{9}, \infty\right)$
- $[1, \infty]$
- $[0, \infty]$
For both roots to exceed 3, the vertex must be greater than 3, the discriminant must be non-negative, and the function value at x=3 must be positive. Solving -b/(2a) > 3 gives 3m > 3, so m > 1. The discriminant 36m^2 - 4(9m^2 - 2m + 2) >= 0 simplifies to 8m - 8 >= 0, or m >= 1. Finally, f(3) = 9 - 18m + 9m^2 - 2m + 2 = 9m^2 - 20m + 11 > 0, which factors to (9m - 11)(m - 1) > 0, implying m > 11/9 or m < 1.
For both roots to exceed 3, the quadratic must have real roots, the vertex must be to the right of 3, and the function evaluated at 3 must be positive. First, ensure the discriminant is non-negative: (negative 6m) squared minus 4 times 1 times (9m squared minus 2m plus 2) simplifies to 8m minus 8, requiring m to be greater than or equal to 1. Next, the axis of symmetry, given by 3m, must be greater than 3, which also requires m to be greater than 1. Finally, substituting x equals 3 gives 9 minus 18m plus 9m squared minus 2m plus 2, which simplifies to 9m squared minus 20m plus 11; setting this greater than zero yields (9m minus 11)(m minus 1) greater than zero, meaning m must be strictly less than 1 or strictly greater than 11 divided by 9. Intersecting the condition m greater than 1 with this result leaves m strictly greater than 11 divided by 9, giving the interval (11 divided by 9, infinity).