slide 2- 1€¦ · slide 2- 10 extraneous solutions when we multiply or divide an equation by an...

Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

2.8 Solving Equations in One Variable


Quick Review

2

2

Find the missing numerator or denominator.2 ?1.

3 2 34 162.4 ?

Find the LCD and rewrite the expression as a single fractionreduced to lowest terms.

3 5 73. 2 4 12

24. 1

Use the quadratic for

x x xx xx

xx x

=+ + −− −

=+

+ −

+−

2

mula to find the zeros of the quadratic polynomial.

5. 2 4 1x x+ −


Quick Review Solutions

2

22

Find the missing numerator or denominator.2 ?1.

3 2 34 162. 4 ?

Find the LCD and rewrite the expression as a single fractionreduced to lowest ter

2 2

8 1

ms.3 5 73. 2

6

14

3 / 612

x x xx x xx

x

x

−

+ +

=+ + −− −

=+

+ −

2

2

2

24. 1

Use the quadratic formula to find the zeros of the quadratic polynomial.

5. 2 4 1

2 2

2 62

x x xx

x x

x xx+

− −

+

−

− ±−

+


What you’ll learn about

Solving Rational Equations Extraneous Solutions Applications … and why Applications involving rational functions as models often require that an equation involving fractions be solved.


Example Solving by Clearing Fractions

2Solve 3.xx

+ =


Example Solving by Clearing Fractions

2Solve 3.xx

+ =

2

2

The LCD is .2 3

2 3

3 2 0 ( 2)( 1) 0

2 or

multiply by

subtract 3factor

use zero factor prop 1Confirm algebraically:

2Let 2 :

erty

E

2 32

2Let 1: 1 3 1

x

xx

x xx x

xx

x

x

xx x

x

+ =

+ =

− + =− − == =

= + =

= + = ach value is a solution of the original equation.


Solving a Rational Equation

Slide 2- 8

1Solve 0.4

xx

+ =−


Solve Graphically The graph in Figure 2.56 suggests that the function has two zeros. We can use the graph to find that the zeros are about 0.268 and 3.732, agreeing with the values found algebraically.

Solving a Rational Equation

Slide 2- 9

1x+ = 0.x - 4


Extraneous Solutions When we multiply or divide an equation by an expression containing variables, the resulting equation may have solutions that are not solutions of the original equation. These are extraneous solutions. For this reason we must check each solution of the resulting equation in the original equation.


Example Eliminating Extraneous Solutions

21 2 2Solve the equation .

3 1 4 3x

x x x x+ =

− − − +



21 2 2Solve the equation .

3 1 4 3x

x x x x+ =

− − − +

2

2

The LCD is ( 1)( 3).1 2 2( 1)( 3) ( 1)( 3)

3 1 4 3( 1)(1) 2 ( 3) 2

2 5 3 0(2 1)( 3) 0

1/ 2 or 3 in the original equation. 1 / 2 is the

only solutio

x xxx x x x

x x x xx x xx xx x

x xx

− −

− − + = − − − − − + − + − =

− − =+ − =

= − == −Check solutions

n. The original equation is not defined at 3.x =



23 3 6Solve the equation 0

2 2x

x x x x−

+ + =+ +


Calculating Acid Mixtures

Slide 2- 14

How much pure acid must be added to 50 mL of a 35% acid solution to produce a mixture that is 75% acid?


Calculating Acid Mixtures

Slide 2- 15

How much pure acid must be added to 50 mL of a 35% acid solution to produce a mixture that is 75% acid?


.

.35(50) .75(50 )

80

OR

set up the following equation:

Let x theamount of acid tobeadded

x x

x

=

+ = +

=

acid in the current solution + acid added = acid in the final solution


Example Finding a Minimum Perimeter

Find the dimensions of the rectangle with minimum perimeter if its area is 200 square meters. Find this least perimeter.


Example Finding a Minimum Perimeter

Find the dimensions of the rectangle with minimum perimeter if its area is 200 square meters. Find this least perimeter.

Word Statement: Perimeter 2 length 2 width width in meters 200 / length in meters

200 400Function to be minimized: ( ) 2 2 2

x x

P x x xx x

= × + ×= =

= + = +

Solve graphically: A minimum of approximately 56.57 occurs when 14.14The width is 14.14 m and the length is 200/14.14=14.14 m,a square.The minimum perimeter is 56.57 m.

x ≈


Designing a Juice Can

Slide 2- 21

Stewart Cannery will package tomato juice in 2-liter cylindrical cans. Find the radius and height of the cans if the cans have a surface area of 1000 cm2. (1 liter = 1000 cm3)

Model S = surface area of can in cm2

r = radius of can in centimeters h = height of can in centimeters


3

2 2

Using volume V and surface area S formulas and the fact that 1 L = 1000 cm , we conclude that

V = r h = 2000 and S = 2 r + 2 rh = 1000.π π π



pg 254: ,#1-35 odd

slide 2- 1€¦ · slide 2- 10 extraneous solutions when we multiply or divide an equation by an...

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