2. The Discriminant It is sometimes enough to know what type of number a solution will be, without actually solving the equation. From the quadratic formula, b 2 – 4ac, is known as the discriminant. The discriminant determines what type of number the solutions of a quadratic equation are. The cases are summarized on the next slide.

4. Example Solution For the equation 4x 2 – x + 1 = 0, determine what type of number the solutions are and how many exist. First determine a, b, and c: a = 4, b = –1, and c = 1. Compute the discriminant: b 2 – 4ac = (–1) 2 – 4(4)(1) = –15. Since the discriminant is negative, there are two imaginary-number solutions that are complex conjugates of each other.

5. Example Solution For the equation 5x 2 – 10x + 5 = 0, determine what type of number the solutions are and how many exist. First determine a, b, and c: a = 5, b = –10, and c = 5. Compute the discriminant: b 2 – 4ac = (–10) 2 – 4(5)(5) = 0. There is exactly one solution, and it is rational. This indicates that 5x 2 – 10x + 5 = 0 can be solved by factoring.

6. Example Solution For the equation 2x 2 + 7x – 3 = 0, determine what type of number the solutions are and how many exist. First determine a, b, and c: a = 2, b = 7, and c = –3. Compute the discriminant: b 2 – 4ac = (7) 2 – 4(2)(–3) = 73. The discriminant is a positive number that is not a perfect square. Thus there are two irrational solutions that are conjugates of each other.

7. Writing Equations from Solutions We know by the principle of zero products that (x – 1)(x + 4) = 0 has solutions 1 and  4. If we know the solutions of an equation, we can write an equation, using the principle in reverse.

8. Example Solution Find an equation for which 5 and –4/3 are solutions. x = 5 or x = –4/3 x – 5 = 0 or x + 4/3 = 0 (x – 5)(x + 4/3) = 0 x 2 – 5x + 4/3x – 20/3 = 0 3x 2 – 11x – 20 = 0 Get 0’s on one side Using the principle of zero products Multiplying Combining like terms and clearing fractions

9. Example Solution Find an equation for which 3i and –3i are solutions. x = 3i or x = –3i x – 3i = 0 or x + 3i = 0 (x – 3i)(x + 3i) = 0 x 2 – 3ix + 3ix – 9i 2 = 0 x 2 + 9 = 0 Get 0’s on one side Using the principle of zero products Multiplying Combining like terms

10. Example Find an equation for which – 4, 0 and 1 are solutions. Solution