53 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, \( h(x)=f(x)-3 \). Given: \( (x+5)^{2}=1-p^{2} \) Calculate the values of \( p \) for which the roots of the equation are non-real.

To find the coordinates of the turning point of the function \( h(x) = f(x) - 3 \), we first need to analyze the function \( f(x) = x^{2} - p x + 8 + 2p \) with the condition of equal roots. The condition for equal roots is given by the discriminant being zero: \[ D = (-p)^2 - 4(1)(8 + 2p) = 0 \] This simplifies to: \[ p^2 - 32 - 8p = 0 \implies p^2 + 8p - 32 = 0 \] Using the quadratic formula \( p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): \[ p = \frac{-8 \pm \sqrt{8^2 + 4 \times 32}}{2} = \frac{-8 \pm \sqrt{64 + 128}}{2} = \frac{-8 \pm 12}{2} \] This gives us: \[ p = 2 \text{ or } p = -10 \] Given that \( p < 0 \), we take \( p = -10 \). Now, substituting \( p \) into \( f(x) \): \[ f(x) = x^{2} + 10x + 8 - 20 = x^{2} + 10x - 12 \] The vertex (turning point) of a parabola \( ax^2 + bx + c \) is given by \( x = -\frac{b}{2a} \): \[ x = -\frac{10}{2} = -5 \] To find the y-coordinate of the turning point \( f(-5) \): \[ f(-5) = (-5)^{2} + 10(-5) - 12 = 25 - 50 - 12 = -37 \] Thus, the turning point of \( h(x) = f(x) - 3 \) becomes: \[ h(-5) = f(-5) - 3 = -37 - 3 = -40 \] So, the coordinates of the turning point are \( (-5, -40) \). Next, for the condition \( (x + 5)^{2} = 1 - p^{2} \) to yield non-real roots, the right side must be negative: \[ 1 - p^{2} < 0 \implies p^{2} > 1 \] The solutions to this inequality are: \[ p < -1 \text{ or } p > 1 \] Since we are told that \( p < 0 \), the condition simplifies to: \[ p < -1 \] Thus, the values of \( p \) for which the roots of the equation are non-real are \( p < -1 \).