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4.7 Quadratic Equations and Problem Solving BobsMathClass.Com Copyright © 2010 All Rights Reserved. 1 General Strategy for Problem Solving Understand the problem Read and reread the problem Choose a variable to represent the unknown Construct a drawing, whenever possible Propose a solution and check Translate the problem into an equation Solve the equation Interpret the result Check proposed solution in problem State your conclusion

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4.7 Quadratic Equations and Problem Solving BobsMathClass.Com Copyright © 2010 All Rights Reserved. 2 Solving Applications using factoring and the Principle of Zero Products We have covered similar word problems to ones we are about to solve. However, we now have more methods for solving equations. To receive full credit for a word problem on a test, all steps must be shown. If you find the correct answer without the correct equation and necessary steps, you will still receive no credit. So show the steps. Example 1. Find two consecutive integers whose product is 56. Solution: x = the first integer x+1 = the second integer (consecutive) (Since the product is 56.) x is equal to the first integer. There are two different choices for x. If x = –8, then the second integer is –8+1 = –7. If x = 7, then the second integer is 7+1 = 8. Answer: – 8 & – 7, or 7 & 8 1. Define the unknowns. 2. State the equation and solve. 3. Answer the question. Your Turn Problem #1 Find two consecutive even integers whose product is 168.

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4.7 Quadratic Equations and Problem Solving BobsMathClass.Com Copyright © 2010 All Rights Reserved. 3 Example 2. The sum of the squares of two consecutive odd integers is 130. Solution: Let x = the first integer Let x+2 = the second integer (consecutive odd) Sum of squares equals 130. There are two different choices for x. If x = –9, the second integer is –9+2 = –7. If x = 7, the second integer is 7+2 = 9. Your Turn Problem #2 The sum of the square of two consecutive integers is 181. Find the integers.

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4.7 Quadratic Equations and Problem Solving BobsMathClass.Com Copyright © 2010 All Rights Reserved. 4 x 2 + (x + 10) 2 = (2x – 10) 2 x 2 + x 2 + 20x + 100 = 4x 2 – 40x + 100 x = 0 or x = 30 0 = 2x(x – 30) 0 = 2x 2 – 60x Let x = the length of the shorter leg, then x + 10 = the length of the longer leg and 2x – 10 = the length of the hypotenuse. Example 3: Solution: Remember that x represented the length of the shorter side. So, although x = 0 satisfies our equation, it cannot be a solution for the problem we were presented. If we let x = 30, then x + 10 = 40 and 2x – 10 = 50 Answer: Shorted leg = 30, longer leg = 40, hypotenuse = 50 Find the length of each side of the right triangle if the longer leg is 10 miles more than the shorter leg and the hypotenuse is 10 miles less than twice the shorter leg. Your Turn Problem #3 The length of a rectangle is 1 ft less than twice the width. Its area is 120 ft 2. Find the dimensions. Using the Pythagorean Theorem, Answer: width = 8 ft, length =15 ft.

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4.7 Quadratic Equations and Problem Solving BobsMathClass.Com Copyright © 2010 All Rights Reserved. 5 When the object hits the ground, we will have h= 0, so we wish to solve the equation -16t 2 + 80t + 576 = 0. So the object hits the ground 9 seconds after it is released. Example 4. An object is thrown upward with an initial velocity of 80 feet per second off the top of a 576 foot tall building. The height h in feet of the object after t seconds is given by the quadratic equation h=-16t 2 + 80t + 576. When will the object hit the ground. Solution: We can factor a -16 from every term to give -16(t 2 - 5t – 36) = 0, which factorizes to -16(t - 9)(t + 4) = 0. Discarding the negative root, we find that t = 9. Your Turn Problem #4 Answer: 5 seconds

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