Given \( f(x)=3(x-1)^{2}+5 \) and \( g(x)=3 \) Determine the values of \( k \) for which \( q(a)=g(x)+k \) has Two unequal real roots.

To determine the values of \( k \) for which the equation \( q(x) = g(x) + k = 3 + k \) has two unequal real roots, we start by transforming the context a bit. Here \( q(x) \) is a constant function because \( g(x) = 3 \). Now, since \( q(x) \) is a constant, it cannot have any roots unless it is equal to a quadratic function that can potentially cross the x-axis. The function \( f(x) = 3(x-1)^2 + 5 \) is a parabola that opens upwards. The vertex of this parabola is at \( x = 1 \) with a minimum value of \( f(1) = 5 \). For \( q(x) \) to intersect \( f(x) \) and have two unequal real roots, we require that \( q(x) = 3+k \) must be less than the minimum value of \( f(x) \). Thus, we set up the inequality: \[ 3 + k < 5 \] Solving this gives: \[ k < 2 \] Therefore, for \( k < 2 \), the line \( y = q(x) = 3 + k \) intersects the parabola \( f(x) \) at two points, resulting in two unequal real roots. In conclusion, the values of \( k \) for which \( q(x) = g(x) + k \) has two unequal real roots are: \[ k < 2 \]