Presentation is loading. Please wait.

Presentation is loading. Please wait.

WARM UP Solve 1. 2. or. USING QUADRATIC EQUATIONS.

Similar presentations


Presentation on theme: "WARM UP Solve 1. 2. or. USING QUADRATIC EQUATIONS."— Presentation transcript:

1 WARM UP Solve 1. 2. or

2 USING QUADRATIC EQUATIONS

3 SOLVING PROBLEMS For some problems a quadratic equation will serve as a mathematical model. Problem-solving strategies such as write an equation, draw a diagram, and others may be used together, as with a linear model. PROBLEM SOLVING GUIDELINES UUNDERSTAND the problem DDevelop and carry out a PLAN FFind the ANSWER and CHECK

4 EXAMPLE A rectangular lawn is 60m by 80m. Part of the lawn is torn up to install a pool, leaving a strip of lawn of uniform width around the pool. The area of the pool is 1/6 of the old lawn area. How wide is the strip of lawn? UUNDERSTAND the problem Question: Find the width of the strip of lawn Data: The lawn is 60m by 80m The pool is 1/6 of the total area.

5 EXAMPLE CONT. DDevelop and carry out a PLAN First draw a diagram 60 60 – 2x 80 Total area = 60  80 Area of pool = (60 – 2x)(80 – 2x) The area of the pool is 1/6 of original area of the lawn. Therefore (60 – 2x)(80 – 2x) = 1/6  60  80 4800 – 160x – 120x + 4x = 800 4x – 280x + 4000 = 0 x – 70x + 1000 = 0 (x – 20)(x – 50) = 0 x = 20 or x = 50

6 EXAMPLE CONT. FFind the ANSWER and CHECK We see that 50 cannot be a solution because when x is 50, 60 – 2x, which is the width of the pool, is -40 But the width of the pool cannot be negative. A 20 meter wide strip checks in the problem. Since the width must be smaller than 60, it is a reasonable answer.

7 TRY THIS… An open box is to be made from a 10 cm by 20 cm rectangular piece of cardboard by cutting a square from each corner. The area of the bottom of the box is 96 cm. What is the length of the sides of the squares that are cut from the corners?

8 THEOREM 8-1 The following theorem is helpful for solving problems involving quadratic equations. The Pythagorean Theorem In any right triangle, if a and b are the lengths of the legs and c is the length of the hypotenuse, then

9 EXAMPLE Bicycles A and B leave the same point P at the same time at right angles. B travels 7 km/h faster than A. After 3 hours they are 39 km apart. Find the speed of each. We first make a drawing, letting r be the speed of A and r + 7 be the speed of of B. Since they both travel 3 hours, their distances from P are 3r and 3(r + 7). Using the Pythagorean Theorem Multiplying by 1/9 Finding standard form Multiplying by 1/2 The solutions of the equations are -12 and 5. Since speed cannot be negative, -12 is not a solution. The speed of A is 5 km/h and the speed of B is 12 km/h.

10 TRY THIS… Runners A and B leave the same point P at right angles. A runs 4 km/h faster than B. After 2 hours they are 40 km apart. Find the speed of each.

11 CH. 8.2 HOMEWORK Textbook pg. 349 #2, 6, 8 & 10


Download ppt "WARM UP Solve 1. 2. or. USING QUADRATIC EQUATIONS."

Similar presentations


Ads by Google