College Algebra, Chapter 1, 1.3, Section 1.3, Problem 4

Make up quadratic equations that have the following number of solutions:

a.) Two Solutions

For two solutions, the discriminant $b^2 - 4ac > 0$, so we let

$x^2 - 5x - 4 = 0$

b.) One solutions

For one solution, the discriminant $b^2 - 4ac = 0$, so we let

$x^2 - 8x + 16 = 0$

c.) No solution

For no solution, the discriminant $b^2 - 4ac < 0$, so we let

$x^2 - 3x + 8 = 0$