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Quadratic Equations: Solving Real-Life Problems

Learn how to solve real-life problems modeled by quadratic equations, with examples including cliff diving and geometric calculations.

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Quadratic Equations: Solving Real-Life Problems

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  1. 6 Chapter Chapter 2 Factoring Polynomials

  2. Quadratic Equations and Problem Solving Section 6.7

  3. Solving Problems Modeled by Quadratic Equations Objective 1

  4. Example: Free Fall • Cliff divers frequent the falls at Waimea Falls Park in Oahu, Hawaii. One of the popular diving spots is 64 feet high. Neglecting air resistance, the height of a diver above the pool after t seconds is h = –16t2 + 64. Find how long it takes a diver to reach the pool.

  5. Example (cont) • 1. UNDERSTAND. Read and reread the problem. The equation h = –16t2 + 64 models the height of the falling diver at time t. Familiarize yourself with this equation by finding the height of the diver at time t = 1 second. • When t = 1 second, the height of the diver is h = –16(1)2 + 64 = 48 feet.

  6. Example (cont) • 2 and 3 Translate and Solve.

  7. Example (cont) • 4 Interpret • Since the time t cannot be negative, the proposed solution is 2 seconds. • Check: Verify that the height of the diver when t is 2 seconds is 0. • h = –16(2)2 + 64 = 0 feet. • State: It takes the diver 2 seconds to reach the ocean.

  8. Example • The square of a number minus twice the number is 63. Find the number. • Translate into an equation: x2 – 2x = 63 • Solve.

  9. Example • The length of a rectangular garden is 5 feet more than its width. The area of the garden is 176 square feet. Find the length and the width of the garden. • Let x = the width • Let x + 5 = length • A = length times width • 176 = x(x + 5)

  10. Example (cont) • 176 = x(x + 5) Since x represents the length of the base, we discard the solution –16. The width is 11 feet and the length is 16 feet.

  11. 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

  12. The product of is 132 two consecutive positive integers 132 x (x + 1) = • Example (cont) 2. TRANSLATE Continued

  13. Example (cont) 3. SOLVE x(x + 1) = 132 x2 + x = 132 Apply the distributive property. x2 + x – 132 = 0 Write in standard form. (x + 12)(x – 11) = 0 Factor. x + 12 = 0 or x – 11 = 0 Set each factor equal to 0. x = –12 or x = 11 Solve. Continued

  14. Example (cont) 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.

  15. The Pythagorean Theorem • PythagoreanTheorem • 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.

  16. 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. Continued

  17. Example (cont) 2. TRANSLATE By the Pythagorean Theorem, (leg)2 + (leg)2 = (hypotenuse)2 x2 + (x + 10)2 = (2x – 10)2 Continued

  18. Multiply the binomials. x2 + x2 + 20x + 100 = 4x2 – 40x + 100 Combine like terms. 2x2 + 20x + 100 = 4x2 – 40x + 100 Write in standard form. 0 = 2x2 – 60x 0 = 2x(x – 30) Factor. Set each factor equal to 0 and solve. x = 0 or x = 30 Example (cont) 3. SOLVE x2 + (x + 10)2 = (2x – 10)2 Continued

  19. Example (cont) 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.)

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