Find a possible pair of integer values for \( a \) and \( c \) so that the equation \( a x^{2}+8 x+c=0 \) has the given number and type of solution(s). Then write the equation. \( \begin{array}{lll}\text { 17. two real solutions } & \text { 18. two imaginary solutions } & \text { 19. one real solution }\end{array} \)

For the equation \( a x^{2}+8 x+c=0 \): 1. For two real solutions, the discriminant must be greater than zero. The discriminant \( D = b^2 - 4ac = 8^2 - 4ac > 0 \). Let's find \( a = 1 \) and \( c = -5 \). The equation becomes \( 1x^2 + 8x - 5 = 0 \), and the discriminant is \( 64 + 20 = 84 > 0\), giving us two real solutions. 2. For two imaginary solutions, the discriminant must be less than zero. Let's choose \( a = 1 \) and \( c = 10 \). The resulting equation \( 1x^2 + 8x + 10 = 0 \) has a discriminant of \( 64 - 40 = 24 < 0 \), thus yielding two imaginary solutions. 3. For one real solution, the discriminant must be equal to zero. We can pick \( a = 1 \) and \( c = 16 \). Thus the equation is \( 1x^2 + 8x + 16 = 0 \), and the discriminant is exactly \( 64 - 64 = 0\), providing one real solution.