Download presentation

Presentation is loading. Please wait.

Published byCarlie Follen Modified over 2 years ago

1
Copyright © 2010 Pearson Education, Inc. Quadratic Equations and Problem Solving Understand basic concepts about quadratic equationsUnderstand basic concepts about quadratic equations Use factoring, the square root property, completing the square, and the quadratic formula to solve quadratic equationsUse factoring, the square root property, completing the square, and the quadratic formula to solve quadratic equations Understand the discriminantUnderstand the discriminant Solve problems involving quadratic equationsSolve problems involving quadratic equations 3.2

2
Slide Copyright © 2010 Pearson Education, Inc. Quadratic Equation A quadratic equation in one variable is an equation that can be written in the form ax 2 + bx + c where a, b, and c are constants with a 0.

3
Slide Copyright © 2010 Pearson Education, Inc. Solving Quadratic Equations Quadratic equations can have no real solutions, one real solution, or two real solutions. The are four basic symbolic strategies in which quadratic equations can be solved. Factoring Square root property Completing the square Quadratic formula

4
Slide Copyright © 2010 Pearson Education, Inc. Factoring Factoring is a common technique used to solve equations. It is based on the zero- product property, which states that if ab = 0, then a = 0 or b = 0 or both. It is important to remember that this property works only for 0. For example, if ab = 1, then this equation does not imply that either a = 1 or b = 1. For example, a = 1/2 and b = 2 also satisfies ab = 1 and neither a nor b is 1.

5
Slide Copyright © 2010 Pearson Education, Inc. Example 1b Solve the quadratic equation 12t 2 = t + 1. Check your results. Check: Solution

6
Slide Copyright © 2010 Pearson Education, Inc. Example 1b Solution continued Check:

7
Slide Copyright © 2010 Pearson Education, Inc. The Square Root Property Let k be a nonnegative number. Then the solutions to the equation x 2 = k are given by

8
Slide Copyright © 2010 Pearson Education, Inc. Example 4 If a metal ball is dropped 100 feet from a water tower, its height h in feet above the ground after t seconds is given by h(t) = 100 – 16t 2. Determine how long it takes the ball to hit the ground. Solution The ball strikes the ground when the equation 100 – 16t 2 = 0 is satisfied.

9
Slide Copyright © 2010 Pearson Education, Inc. Example 4 Solution continued The ball strikes the ground after 10/4, or 2.5, seconds.

10
Slide Copyright © 2010 Pearson Education, Inc. Completing the Square Another technique that can be used to solve a quadratic equation is completing the square. If a quadratic equation is written in the form x 2 + kx =d, where k and d are constants, then the equation can be solved using

11
Slide Copyright © 2010 Pearson Education, Inc. Example Solve 2x 2 – 8x = 7. Solution Divide each side by 2:

12
Slide Copyright © 2010 Pearson Education, Inc. Symbolic, Numerical and Graphical Solutions Quadratic equations can be solved symbolically, numerically, and graphically. The following example illustrates each technique for the equation x(x – 2) = 3.

13
Slide Copyright © 2010 Pearson Education, Inc. Symbolic Solution

14
Slide Copyright © 2010 Pearson Education, Inc. Numerical Solution

15
Slide Copyright © 2010 Pearson Education, Inc. Graphical Solution

16
Slide Copyright © 2010 Pearson Education, Inc. Quadratic Formula The solutions to the quadratic equation ax 2 + bx + c = 0, where a 0, are given by

17
Slide Copyright © 2010 Pearson Education, Inc. Example 7 Solve the equation 3x 2 – 6x + 2 = 0. Solution Let a = 3, b = 6, and c = 2.

18
Slide Copyright © 2010 Pearson Education, Inc. The Discriminant If the quadratic equation ax 2 + bx + c = 0 is solved graphically, the parabola y = ax 2 + bx + c can intersect the x-axis zero, one, or two times. Each x-intercept is a real solution to the quadratic equation.

19
Slide Copyright © 2010 Pearson Education, Inc. Quadratic Equations and Discriminant To determine the number of real solutions to ax 2 + bx + c = 0 with a 0, evaluate the discriminant b 2 – 4ac. 1. If b 2 – 4ac > 0, there are two real solutions. 2. If b 2 – 4ac = 0, there is one real solution. 3. If b 2 – 4ac < 0, there are no real solutions.

20
Slide Copyright © 2010 Pearson Education, Inc. Example 9 Use the discriminant to find the number of solutions to 9x 2 – 12.6x = 0. Then solve the equation by using the quadratic formula. Support your answer graphically. Solution Let a = 9, b = –12.6, and c = 4.41 b 2 – 4ac = (–12.6) 2 – 4(9)(4.41) = 0 Discriminant is 0, there is one solution.

21
Slide Copyright © 2010 Pearson Education, Inc. Example 9 Solution continued The only solution is 0.7.

22
Slide Copyright © 2010 Pearson Education, Inc. Example 9 Solution continued The graph suggests there is only one intercept 0.7.

23
Slide Copyright © 2010 Pearson Education, Inc. Problem Solving Many types of applications involve quadratic equations. To solve these problems, we use the steps for Solving Application Problems from Section 2.3 on page 122.

24
Slide Copyright © 2010 Pearson Education, Inc. Example 11 A box is being constructed by cutting 2-inch squares from the corners of a rectangular piece of cardboard that is 6 inches longer than it is wide. If the box is to have a volume of 224 cubic inches, find the dimensions of the piece of cardboard.

25
Slide Copyright © 2010 Pearson Education, Inc. Example 11 Solution Step 1:Let x be the width and x + 6 be the length. Step 2:Draw a picture.

26
Slide Copyright © 2010 Pearson Education, Inc. Solution continued Since the height times the width times the length must equal the volume, or 224 cubic inches, the following can be written 2(x – 4)(x + 2) = 224 Step 3: Write the quadratic equation in the form ax 2 + bx + c = 0 and factor.

27
Slide Copyright © 2010 Pearson Education, Inc. Solution continued The dimensions can not be negative, so the width is 12 inches and the length is 6 inches more, or 18 inches. Step 4:After the 2-inch-square corners are cut out, the dimensions of the bottom of the box are 12 – 4 = 8 inches by 18 – 4 = 14 inches. The volume of the box is then 2814 = 224 cubic inches, which checks.

Similar presentations

© 2016 SlidePlayer.com Inc.

All rights reserved.

Ads by Google