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MAT 150 Unit 2-2: Solving Quadratic Equations. Objectives  Solve quadratic equations using factoring  Solve quadratic equations graphically using the.

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Presentation on theme: "MAT 150 Unit 2-2: Solving Quadratic Equations. Objectives  Solve quadratic equations using factoring  Solve quadratic equations graphically using the."— Presentation transcript:

1 MAT 150 Unit 2-2: Solving Quadratic Equations

2 Objectives  Solve quadratic equations using factoring  Solve quadratic equations graphically using the x-intercept method and the intersection method  Solve quadratic equations by combining graphical and factoring methods  Solve quadratic equations using the square root method  Solve quadratic equations by completing the square  Solve quadratic equations using the quadratic formula  Solve quadratic equations having complex solutions

3 Factoring Methods An equation that can be written in the form ax 2 + bx + c = 0, with a ≠ 0, is called a quadratic equation. Zero Product Property If the product of two real numbers is 0, then at least one of them must be 0. That is, for real numbers a and b, if the product ab = 0, then either a = 0 or b = 0 or both a and b are equal to 0.

4 Solve with Factoring

5 Example The height above ground of a ball thrown upward at 64 feet per second from the top of an 80-foot-high building is modeled by S(t) = 80 + 64t – 16t 2 feet, where t is the number of seconds after the ball is thrown. How long will the ball be in the air? Solution

6 Example Consider the daily profit from the production and sale of x units of a product, given by P(x) = –0.01x 2 + 20x – 500 dollars. a. Use a graph to find the levels of production and sales that give a daily profit of $1400. b. Is it possible for the profit to be greater than $1400?

7 Example Consider the daily profit from the production and sale of x units of a product, given by P(x) = –0.01x 2 + 20x – 500 dollars. a. Use a graph to find the levels of production and sales that give a daily profit of $1400. Solution

8 Example (cont) Consider the daily profit from the production and sale of x units of a product, given by P(x) = –0.01x 2 + 20x – 500 dollars. b. Is it possible for the profit to be greater than $1400? Solution

9 Combining Graphs and Factoring Factor Theorem The polynomial function f has a factor (x – a) if and only if f(a) = 0. Thus, (x – a) is a factor of f(x) if and only if x = a is a solution to f (x) = 0.

10 The Square Root Method Square Root Method The solutions of the quadratic equation x 2 = C are x = Note that, when we take the square root of both sides, we use a ± symbol because there are both a positive and a negative value that, when squared, give C.

11 Example Solve the following equations using the square root method. a. 3x 2 – 6 = 0b. (x – 2) 2 = 7 Solution

12 Quadratic Formula The solutions of the quadratic equation ax 2 + bx + c = 0 are given by the formula Note that a is the coefficient of x 2, b is the coefficient of x, and c is the constant term.

13 Example Solve 5x 2 – 8x = 3 using the quadratic formula. Solution

14 The Discriminant We can also determine the type of solutions a quadratic equation has by looking at the expression b 2  4ac, which is called the discriminant of the quadratic equation ax 2 + bx + c = 0. The discriminant is the expression inside the radical in the quadratic formula, so it determines if the quantity inside the radical is positive, zero, or negative. If b 2  4ac > 0, there are two different real solutions. If b 2  4ac = 0, there is one real solution. If b 2  4ac < 0, there is no real solution.

15 Aids for Solving Quadratic Equations

16 Example Solve the equations. a. x 2 = – 36b. 3x 2 + 36 = 0 Solution

17 Example Solve the equations. a. x 2 – 3x + 5 = 0b. 3x 2 + 4x = –3 Solution a.

18 Example (cont) Solve the equations. a. x 2 – 3x + 5 = 0b. 3x 2 + 4x = –3 Solution b.

19 Market Equilibrium


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