## Presentation on theme: "Section 8.4 Quadratic Formula."— Presentation transcript:

Objectives Solving Quadratic Equations The Discriminant

QUADRATIC FORMULA The solutions to ax2 + bx + c = 0 with a ≠ 0 are given by

Example Solve the equation 4x2 + 3x – 8 = 0. Support your results graphically. Solution Symbolic Solution Let a = 4, b = 3 and c = − 8. or or

Example (cont) 4x2 + 3x – 8 = 0 Graphical Solution

Example Solve the equation 3x2 − 6x + 3 = 0. Support your result graphically. Solution Let a = 3, b = −6 and c = 3.

Example Solve the equation 2x2 + 4x + 5 = 0. Support your result graphically. Solution Let a = 2, b = 4 and c = 5. There are no real solutions for this equation because is not a real number.

EQUATIONS To determine the number of solutions to the quadratic equation ax2 + bx + c = 0, evaluate the discriminant b2 – 4ac. 1. If b2 – 4ac > 0, there are two real solutions. 2. If b2 – 4ac = 0, there is one real solution. 3. If b2 – 4ac < 0, there are no real solutions; there are two complex solutions.

Example Use the discriminant to determine the number of solutions to −2x2 + 5x = 3. Then solve the equation using the quadratic formula. Solution −2x2 + 5x − 3 = 0 Let a = −2, b = 5 and c = −3. b2 – 4ac = (5)2 – 4(−2)(−3) = 1 or Thus, there are two solutions.

THE EQUATION x2 + k = 0 If k > 0, the solution to x2 + k = 0 are given by

Example Solve x = 0. Solution The solutions are

Example Solve 3x2 – 7x + 5 = 0. Write your answer in standard form: a + bi. Solution Let a = 3, b = −7 and c = 5. and

Example Solve Write your answer in standard form: a + bi. Solution Begin by adding 2x to each side of the equation and then multiply by 5 to clear fractions. Let a = −2, b = 10 and c = −15.

Example (cont) Let a = −2, b = 10 and c = −15.

Example Solve by completing the square. Solution After applying the distributive property, the equation becomes Since b = −4 ,add to each side of the equation. The solutions are 2 + i and 2 − i.