2. The Quadratic Formula For any quadratic equation of the form The solutions are given by the formula:

3. Solution Solve 3x 2 + 5x = 2 using the quadratic formula. First determine a, b, and c: 3x 2 + 5x – 2 = 0; a = 3, b = 5, and c = –2. Substituting Example

4. The solutions are 1/3 and –2.

5. Solution First determine a, b, and c: x 2 – 2x + 7 = 0; a = 1, b = –2, and c = 7. Solve x 2 + 7 = 2x using the quadratic formula. Substituting Example

6. The solutions are

7. Examples Solve the following quadratic equations:

8. Discriminant The radicand in the quadratic formula is called the discriminant. b 2 – 4ac If the discriminant is positive, then there will be two real solutions If the discriminant is 0, then there will be one real solution. If the discriminant is negative, then there will be two imaginary solutions.

9. Solving Quadratic Equations Method 1: Factoring Method 2: Using Square Root Property Method 3: Completing the Square Method 4: Using Quadratic Formula

10. 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.

11. 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 Example

12. 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 Example

13. The braking distance, d (in feet), of a car going v miles per hour is given by How fast would a car be traveling if its braking distance is 150 ft?