1. Quadratic Equations and Problem Solving

2. Martin-Gay, Developmental Mathematics 2 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

3. Martin-Gay, Developmental Mathematics 3 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. Finding an Unknown Number Example Continued

4. Martin-Gay, Developmental Mathematics 4 Finding an Unknown Number Example continued 2.) Translate Continued two consecutive positive integers x(x + 1) is = 132 The product of

5. Martin-Gay, Developmental Mathematics 5 Finding an Unknown Number Example continued 3.) Solve Continued x(x + 1) = 132 x 2 + x = 132 (Distributive property) x 2 + x – 132 = 0 (Write quadratic in standard form) (x + 12)(x – 11) = 0 (Factor quadratic polynomial) x + 12 = 0 or x – 11 = 0 (Set factors equal to 0) x = – 12 or x = 11 (Solve each factor for x)

6. Martin-Gay, Developmental Mathematics 6 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.

7. Martin-Gay, Developmental Mathematics 7 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

8. Martin-Gay, Developmental Mathematics 8 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. The Pythagorean Theorem 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.

9. Martin-Gay, Developmental Mathematics 9 The Pythagorean Theorem Example continued 2.) Translate Continued 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 2 + 20x + 100 = 4x 2 – 40x + 100(multiply the binomials) 2x 2 + 20x + 100 = 4x 2 – 40x + 100(simplify left side) x = 0 or x = 30(set each factor = 0 and solve) 0 = 2x(x – 30)(factor right side) 0 = 2x 2 – 60x(subtract 2x 2 + 20x + 100 from both sides)

10. Martin-Gay, Developmental Mathematics 10 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.)