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Chapter 5

Polynomials and Factoring

5-2Copyright © 2014, 2010, and 2006 Pearson Education, Inc.

Solving Applications

• Applications

• The Pythagorean Theorem

5.8

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Example

The Mitchell’s are designing a garden. The garden will be in the shape of a rectangle and have an area of 270 square feet. The width of the garden is 3 feet less than the length. Find the length and width.Solution1. Familiarize. We first make a drawing. Recall that the area of a rectangle is Length Width. We let x = the length, in feet. The width is then x 3.

x

x 3

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Example

2. Translate. Rewording: The area of the rectangle is 270 ft2.

Translating: x(x 3) = 270

3. Carry out. We solve the equation. x(x 3) = 270 x2 3x = 270

x2 3x 270 = 0 (x 18)(x + 15) = 0 x 18 = 0 or x + 15 = 0 x = 18 or x = 15

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Example

4. Check. The solutions of the equation are 18 and 15. Since the length must be positive, 15 cannot be a solution.

To check 18, we note that if the length is 18, then the width is x 3 or 15 and the area is 18 ft 15 ft = 270 ft2. Thus the solution checks.

5. State. The garden is 18 feet long and 15 feet wide.

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Example

h

h 6

The math club is designing a brochure. The design calls for a triangle to be placed on the front. The triangle has a base that is 6 centimeters less than the height. If the area of the triangle is 216 cm2. Find the height and base. Solution1. Familiarize. We first make a drawing. The formula for the area ofa triangle is A = ½ (base)(height).We let h = the height, in cm, and the base = h 6, in cm.

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Example

2. Translate. Rewording: The area of the triangle is 216 cm2.

Translating: = 216

3. Carry out. We solve the equation.

1( 6)

2h h

1( 6) 216

2h h

( 6) 432h h 2 6 432h h

2 6 432 0h h ( 24)( 18) 0h h

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Example

3. Carry out. h 24 = 0 or h + 18 = 0 h = 24 or h = 18

4. Check. The height must be positive, so 18 cannot be a solution. Suppose the height is 24 cm. The base would be 24 6, or 18 cm, and the area ½(24)(18), or 216 cm2. These numbers check in the original problem.

5. State. The height of the triangle would be 24 cm and the base would be 18 cm.

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The Pythagorean TheoremIn any right triangle, if a and b are the lengths

of the legs and c is the length of the hypotenuse, then

a2 + b2 = c2 or(Leg)2 + (Other leg)2 = (Hypotenuse)2.

The equation a2 + b2 = c2 is called the Pythagorean equation.*

*The converse of the Pythagorean theorem is also true. That is, if a2 + b2 = c2, then the triangle is a right triangle.

c

b

a

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Example

A 13-ft ladder is leaning against a house. The distance from the bottom of the ladder to the house is 7 ft less than the distance from the top of the ladder to the ground. How far is the bottom of the ladder from the house? Solution1. Familiarize. We first make a drawing. The ladder and the missingdimensions form a right triangle. x = distance from top of the ladder tothe ground x 7 = distance from bottom ladder to house. The hypotenuse has length 13 ft.

13 ft

x

x 7

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Example

2. Translate. Since a right triangle is formed, we can use the Pythagorean theorem:

3. Carry out. We solve the equation.

2 2 2a b c 2 2 27( 13)x x

22 14 49 16( ) 9x x x 22 14 49 169x x

22 14 120 0x x

22 7 60 0x x 2( 12)( 5) 0x x

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Example

3. Carry out. x 12 = 0 or x + 5 = 0 x = 12 or x = 5

4. Check. The integer 5 cannot be a length of a side because it is negative. When x = 12, x 7 = 5, and 122 + 52 = 132. So 12 checks.

5. State. The distance from the bottom of the ladder to the house is 5 ft. The distance from the top of the ladder to the ground is 12 ft.