Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
4.7 Solving Quadratic Functions by Factoring Lesson Objectives: Find common factors between numbers Put equations into factored form Solve quadratic equations in factored form Understand the meaning of the factors of a quadratic and how they relate to the graph of a quadratic function Consider This… After the semester is over, you discover that the math department has changed textbooks (again) so the old books are not needed. Your friend Herman and you decide to get creative. You go to the roof of a twelve-story building and look over the edge to the reflecting pool 160 feet below. You drop your book over the edge at the same instant that Herman chucks his book straight down at 48 feet per second (the initial velocity is negative). The quadratic equation h(t) h0 v0 16t 2 models height as a function of time for these situations where h0 represents the initial height and v0 represents the initial velocity. 1. What is the equation that represents the path your book takes? 2. Rewrite this equation in factored form. 3. What is the equation Herman’s book takes? 4. Rewrite this equation in factored form. 5. How long does it take for your book to hit the water? How long does it take Herman’s book to hit the water? (Find the x-intercepts) 6. By how many seconds does his book beat yours into the water? 1 Summarize When we solve the above equations in factored form, we are identifying the roots or x-intercepts of the function. EXAMPLE: Solve (x – 3)(x – 4) = 0. If I multiply two things together and the result is zero, what can I say about those two things? At least one of them must also be zero. That is, the only way to multiply and get zero is to multiply by zero. This is sometimes called "The Zero Factor Property" or "Rule" or "Principle". You cannot make this statement about any other number! You can only make the conclusion about the factors ("one of them must equal zero") if the product itself equals zero. The Zero Factor Principle tells me that at least one of the factors must be equal to zero. Since at least one of the factors must be zero, I'll set them each equal to zero: x – 3 = 0 or x – 4 = 0 This gives me simple linear equations, and they're easy to solve: x = 3 or x = 4 And this is the solution: x = 3, 4 Practice 1) (k + 1)(k − 5) = 0 2) (a + 1)(a + 2) = 0 3) (4k + 5)(k + 1) = 0 4) (2m + 3)(4m + 3) = 0 5) (2k - 5)(-k - 3) = 0 6) (0.5x - 15)(x + 2) = 0 2 Summarize REMEMBER from 4.4, WHEN the “a” value is equal to 1: Standard Form y x 2 bx c Factored Form y (x m)(x n) To get the FACTORED form: 1. Identify the a, b and c values 2. Find the factors or the numbers that MULTIPLY to “c” AND that ADD to “b” _______ + _______ = b _______ _______ = c 3. These are the factors of the equation 4. Plug these factors in for m and n in the factored form of the equation Now to SOLVE A QUADRATIC EQUATION: Solve x2 + 5x + 6 = 0. x2 + 5x + 6 = (x + 2)(x + 3) 1. Set the equation equal to 0. (x + 2)(x + 3) = 0 2. Factor the quadratic equation (notes above) x + 2 = 0 or x + 3 = 0 3. Like before, set each factor equal to 0. x = –2 or x = – 3 4. Solve for x. DON’T FORGET TO CHECK YOUR SOLUTIONS by plugging the values back into the equation for x. The solution to x2 + 5x + 6=0 is x = –3, –2 3 Summarize Solve: Remember: You must have the quadratic equal to zero before you can solve. (x + 2)(x + 3) = 12. You cannot set each of the factors above equal to the other side of the equation and "solve". Instead, you first have to multiply out and simplify the left-hand side, then subtract the 12 over to the left-hand side, and re-factor. Only then can you solve. (x + 2)(x + 3) = 12 x + 5x – 6 = 0 (x + 6)(x – 1) = 0 x2 + 5x + 6 = 12 2 x + 6 = 0 or x – 1 = 0 x = –6 or x = 1 Then the solution to (x + 2)(x + 3) = 12 is x = –6, 1 4 Practice Find the value(s) of x for the following equations. Check using FOIL. 1. x 2 8x 15 0 2. x(x 5) 0 3. x 2 4x 32 0 4. (2x 3)(x 2) 0 5. (4x 3)(2x 7) 0 6. (2x 9)(3x 5) 0 7. x(x 5) 2 8. (x 3)(x 8) 30 9. (x 1)(x 4) 14 10. (x 4)(x 5) 52 11. (x 4)(x 8) 35 12. (x 1)(x 6) 10 13. x 2 12x 35 0 14. x 2 22x 120 0 15. 2x 2 22x 24 0 5 Summarize REMEMBER from 4.4, What happens when “a” is not equal to 1? FIRST: CHECK TO SEE IF YOU CAN FACTOR OUT THE “a” value! If not, then do the following: The Not As Simple Case (a 1) The Box Method can be used to factor quadratics, including the simple case, but it is very useful when a 1. Be sure to have the quadratic in its simplest form. • Factor 4x2 + 4x – 15 5) Determine which sum or difference of these factors will give you the bx term: 1) Make a box and divide it into 4 squares: bx = 4x → -6x+10x = 4x 6) Put these terms in the remaining boxes: 4x2 -6x 2) Put the ax2 term in the top left corner: 10x -15 4x2 7) Factor out common terms to the outside of the box 3) Put the c term in the bottom right corner: 4x 2x 2 5 -15 4) Multiply both terms (ax2 and c) → generate a list of factors. 4x2 • -15 = -60x2 → ± 2x, ± 30x 3x, 20x 4x, 15x 5x, 12x 6x, 10x 2x -3 4x2 -6x 10x -15 8) Read factors from the sides of the box: (2x – 3) (2x + 5), which is the factored form of the quadratic. 6 BE SURE TO CHECK YOUR ANSWER by using FOIL. Practice Solve the following equations. Check your answer using FOIL. 1. 2x 2 6x 4 0 2. 3x 2 15x 18 0 3. x 2 7x 18 0 4. 2x 2 3x 1 0 5. 5x 2 6x 1 0 6. 2x 2 3x 1 0 7. 11x 2 54x 5 0 8. 3x 2 16x 16 0 9. 3x 2 22x 16 0 10. 16b2− 40b + 25 = 0 11. 3b2− 6b + 3 = 0 12. 100x2− 180x + 81 = 0 7 Summarize When we have quadratics in the form Solve x 2 y ax 2 bx : – 5x = 0. Do not divide both sides by x, or you'll lose one of your solutions! This two-term quadratic is easier to factor than were the previous quadratics. FIRST, you can factor an x out of both terms, taking the x out front. x(x – 5) = 0 THEN, set each factor equal to 0 and solve. x = 0 or x – 5 = 0 x = 0 or x = 5 The solution to x2 – 5x = 0 is x = 0, 5 ______________________________________________________________________ When we have quadratics in the form Solve x2 y ax 2 c : – 4 = 0. x2 – 4 = 0 With quadratics in this form, you can isolate the squared variable term, putting the number over on the other side, and solving for x. x2 = 4 When solving an equation, you can do whatever you like to that equation as long as you do the same thing to both sides of the equation. On the lefthand side of this particular equation, you have an x2, and you need a plain x. To turn an x2 into an x, you can take the square root of each side of the equation. I know that: x=±2 Then the solution is x = ± 2 When you square root both sides, remember, the solution gives both the POSITIVE and NEGATIVE value since a negative number squared gives you a positive. 8 Practice Using any of the methods mentioned in this lesson, factor the following expressions. 2. 4x2 1. 16n2 − 9n = 0 4. x2 10. -4x2 13. -16x x2 n2 − 49 = 0 8. 1- x2 7. -36x2 + 400 = 0 9x2 − 1 = 0 6. a4 -4x2 x 14. 2xx2 x2 9 9. -36n2 − 142 = 2 -x2 x 15. -12x x2 x Apply Quadratic functions appear in many different types of word problems. Try these problems by drawing a diagram first, then solving the quadratic equation that corresponds. 1. The length of a rectangle is 6 inches more than its width. The area of the rectangle is 91 square inches. Find the dimensions of the rectangle. 2. The hypotenuse of a right triangle is 6 more than the shorter leg. The longer leg is three more than the shorter leg. Find the length of the shorter leg. 3. One leg of a right triangle is one inch shorter than the other leg. If the hypotenuse is 5 inches, find the length of the shorter leg. 4. The length of a rectangle is 2 less than three times the width. Find the dimensions of the rectangle if the area is 65 square meters. 10