Given \( 2 m x^{2}=3 x-8 \) where \( m \neq 0 \). Determine the value(s) of \( m \) for which the roots of the equation are non-real. Given that \( f(x)=x^{2}-p x+8+2 p \) has two equal root and \( p<0 \). determine the coodinates of the turning point, \( h(x)=f(x)-3 \).

To determine when the roots of the equation \( 2m x^2 = 3x - 8 \) are non-real, we can begin by rewriting it in standard form: \( 2m x^2 - 3x + 8 = 0 \). For the roots to be non-real, the discriminant must be less than zero. The discriminant \( D \) is given by: \[ D = b^2 - 4ac = (-3)^2 - 4(2m)(8) = 9 - 64m \] Setting the discriminant less than zero gives: \[ 9 - 64m < 0 \implies 64m > 9 \implies m > \frac{9}{64}. \] Now, for the function \( f(x) = x^2 - px + 8 + 2p \) to have two equal roots, the discriminant must equal zero: \[ D_f = p^2 - 4(1)(8 + 2p) = 0. \] Expanding this gives: \[ p^2 - 32 - 8p = 0 \implies p^2 + 8p - 32 = 0. \] Using the quadratic formula, we can find \( p \): \[ p = \frac{-8 \pm \sqrt{(8)^2 + 4 \cdot 32}}{2} = \frac{-8 \pm \sqrt{64 + 128}}{2} = \frac{-8 \pm 12}{2}. \] This results in two values: \[ p = 2 \text{ and } p = -10. \] Since \( p < 0 \), we choose \( p = -10 \). Now we substitute \( p = -10 \) back into the function to find the turning point. The turning point occurs at \( x = \frac{p}{2} = \frac{-(-10)}{2} = 5 \). Next, substituting \( x = 5 \) into \( f(x) \): \[ f(5) = 5^2 - (-10)5 + 8 + 2(-10) = 25 + 50 + 8 - 20 = 63. \] So, the coordinates of the turning point are \( (5, 63) \). Finally, to find the turning point of \( h(x) = f(x) - 3 \), we adjust the \( y \)-coordinate by shifting down 3 units. So, the new coordinates will be: \[ (5, 63-3) = (5, 60). \] Thus, the coordinates of the turning point of \( h(x) \) are \( (5, 60) \).