Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall Quadratic Equations and Problem Solving

Martin-Gay, Developmental Mathematics, 2e 22 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Strategy for Problem Solving General Strategy for Problem Solving 1.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 2.TRANSLATE the problem into an equation. 3.SOLVE the equation. 4.INTERPRET the result. Check proposed solution in original problem. State your conclusion.

Martin-Gay, Developmental Mathematics, 2e 33 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Example The product of two consecutive positive integers is 132. Find the two integers. 1. UNDERSTAND Read and reread the problem. If we let x = one of the unknown positive integers, then x + 1 = the next consecutive positive integer. continued

Martin-Gay, Developmental Mathematics, 2e 44 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. continued 2. TRANSLATE continued two consecutive positive integers x(x + 1) is = 132 The product of

Martin-Gay, Developmental Mathematics, 2e 55 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. continued 3. SOLVE continued x(x + 1) = 132 x 2 + x = 132 x 2 + x – 132 = 0 Write in standard form. (x + 12)(x – 11) = 0 x + 12 = 0 or x – 11 = 0 Set each factor equal to 0. x = – 12 or x = 11 Solve. Factor. Apply the distributive property.

Martin-Gay, Developmental Mathematics, 2e 66 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. continued 4.INTERPRET Check: Remember that x should represent a positive integer. So, although x = ‒ 12 satisfies our equation, it cannot be a solution for the problem we were presented. If we let x = 11, then x + 1 = 12. The product of the two numbers is 11 · 12 = 132, our desired result. State: The two positive integers are 11 and 12.

Martin-Gay, Developmental Mathematics, 2e 77 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Pythagorean Theorem In a right triangle, the sum of the squares of the lengths of the two legs is equal to the square of the length of the hypotenuse. (leg a) 2 + (leg b) 2 = (hypotenuse) 2 The Pythagorean Theorem

Martin-Gay, Developmental Mathematics, 2e 88 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Find the length of the shorter leg of a 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. Example continued 1.UNDERSTAND Read and reread the problem. If we 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.

Martin-Gay, Developmental Mathematics, 2e 99 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. 2.TRANSLATE By the Pythagorean Theorem, (leg a) 2 + (leg b) 2 = (hypotenuse) 2 x 2 + (x + 10) 2 = (2x – 10) 2 3.SOLVE x 2 + (x + 10) 2 = (2x – 10) 2 x 2 + x x = 4x 2 – 40x Multiply the binomials. 2x x = 4x 2 – 40x Combine like terms. x = 0 or x = 30 Set each factor equal to 0 and solve. 0 = 2x(x – 30) Factor. 0 = 2x 2 – 60x Write in standard form. continued

Martin-Gay, Developmental Mathematics, 2e 10 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. continued 4.INTERPRET Check: Remember that x is suppose to represent 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. Since = = 2500 = 50 2, the Pythagorean Theorem checks out. State: The length of the shorter leg is 30 miles. (Remember that is all we were asked for in this problem.)