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

Quadratic Equation A quadratic equation in one variable is an equation that can be written in the form ax2 + bx + c where a, b, and c are constants with a ≠ 0. 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 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. Copyright © 2010 Pearson Education, Inc.

Example 1b Solve the quadratic equation 12t2 = t + 1. Check your results. Check: Solution Copyright © 2010 Pearson Education, Inc.

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

The Square Root Property Let k be a nonnegative number. Then the solutions to the equation x2 = k are given by 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 – 16t2. Determine how long it takes the ball to hit the ground. Solution The ball strikes the ground when the equation 100 – 16t2 = 0 is satisfied. Copyright © 2010 Pearson Education, Inc.

Example 4 Solution continued The ball strikes the ground after 10/4, or 2.5, seconds. 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 x2 + kx =d, where k and d are constants, then the equation can be solved using Copyright © 2010 Pearson Education, Inc.

Example Solve 2x2 – 8x = 7. Solution Divide each side by 2: 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. Copyright © 2010 Pearson Education, Inc.

Symbolic Solution Copyright © 2010 Pearson Education, Inc.

Numerical Solution Copyright © 2010 Pearson Education, Inc.

Graphical Solution Copyright © 2010 Pearson Education, Inc.

Quadratic Formula The solutions to the quadratic equation ax2 + bx + c = 0, where a ≠ 0, are given by Copyright © 2010 Pearson Education, Inc.

Example 7 Solve the equation 3x2 – 6x + 2 = 0. Solution Let a = 3, b = 6, and c = 2. Copyright © 2010 Pearson Education, Inc.

The Discriminant If the quadratic equation ax2 + bx + c = 0 is solved graphically, the parabola y = ax2 + bx + c can intersect the x-axis zero, one, or two times. Each x-intercept is a real solution to the quadratic equation. Copyright © 2010 Pearson Education, Inc.

Quadratic Equations and Discriminant To determine the number of real solutions to ax2 + bx + c = 0 with a ≠ 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. Copyright © 2010 Pearson Education, Inc.

Example 9 Use the discriminant to find the number of solutions to 9x2 – 12.6x + 4.41 = 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 b2 – 4ac = (–12.6)2 – 4(9)(4.41) = 0 Discriminant is 0, there is one solution. Copyright © 2010 Pearson Education, Inc.

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

Example 9 Solution continued The graph suggests there is only one intercept 0.7. 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. 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. 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. 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 ax2 + bx + c = 0 and factor. 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 2•8•14 = 224 cubic inches, which checks. Copyright © 2010 Pearson Education, Inc.