University:
California State University, Northridge
Course:
MATH 150A | Calculus I
Academic year:

2024
Views:

288

Pages:

11
Author:

mjmondonedom2x5

Chapter 13 Factoring Polynomials § 13.7 Quadratic Equations and Problem Solving 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 problem • State your conclusion 59 Finding an Unknown Number 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 60 Finding an Unknown Number Example continued 2.) Translate The product of two consecutive positive integers x • (x + 1) is 132 = 132 Continued 61 Finding an Unknown Number Example continued 3.) Solve x(x + 1) = 132 x2 + x = 132 x2 + x – 132 = 0 (x + 12)(x – 11) = 0 x + 12 = 0 or x – 11 = 0 x = –12 or x = 11 (Distributive property) (Write quadratic in standard form) (Factor quadratic polynomial) (Set factors equal to 0) (Solve each factor for x) Continued 62 Finding an Unknown Number Example continued 4.) Interpret Check: Remember that x is suppose to 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. 63 The Pythagorean Theorem 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 leg a hypotenuse leg b 64 The Pythagorean Theorem Example 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. 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. x 2 x - 10 x + 10 Continued 65 The Pythagorean Theorem Example continued 2.) Translate By the Pythagorean Theorem, (leg a)2 + (leg b)2 = (hypotenuse)2 x2 + (x + 10)2 = (2x – 10)2 3.) Solve x2 + (x + 10)2 = (2x – 10)2 x2 + x2 + 20x + 100 = 4x2 – 40x + 100 2x2 + 20x + 100 = 4x2 – 40x + 100 0 = 2x2 – 60x 0 = 2x(x – 30) x = 0 or x = 30 (multiply the binomials) (simplify left side) (subtract 2x2 + 20x + 100 from both sides) (factor right side) (set each factor = 0 and solve) Continued 66 The Pythagorean Theorem Example 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 302 + 402 = 900 + 1600 = 2500 = 502, 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.) 67