Transcript 4.5 Equations, Inequalities, and Applications Involving Root Functions

Chapter 4: Rational, Power, and Root
Functions
4.1 Rational Functions and Graphs
4.2 More on Graphs of Rational Functions
4.3 Rational Equations, Inequalities, Applications,
and Models
4.4 Functions Defined by Powers and Roots
4.5 Equations, Inequalities, and Applications
Involving Root Functions
Copyright © 2007 Pearson Education, Inc.
Slide 4-2
4.5 Equations, Inequalities, and
Applications Involving Root Functions
Power Property
If P and Q are algebraic expressions, then every solution
of the equation P = Q is also a solution of the equation
Pn = Qn, for any positive integer n.
Note: This property does not say that the two
equations are equivalent. The new equation may
have more solutions than the original.
e.g.
x  2
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compared
with
x 4
x  2
2
Slide 4-3
4.5 Solving Equations Involving Root
Functions
1. Isolate a term involving a root on one side of
the equation.
2. Raise both sides of the equation to a power
that will eliminate the radical or rational
exponent.
3. Solve the resulting equation. (If a root is still
present after Step 2, repeat Steps 1 and 2.)
4. Check each proposed solution in the original
equation.
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Slide 4-4
4.5 Solving an Equation Involving Square
Roots
Solve 11  x  x  1 .
Example
Analytic Solution
11  x  x  1
11  x  1  x

11  x   1  x 
2
Isolate the radical.
2
11  x  1  2 x  x
Square both sides.
2
0  x  3 x  10
0  ( x  5 )( x  2 )
x   5 or x  2
2
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Write in standard form
and solve.
Slide 4-5
4.5 Solving an Equation Involving Square
Roots
These solutions must be checked in the original
equation.
Let x   5 .
11  (  5 )  (  5 )  1
16  5  1
9 1
?
?
False. So  5 is an extraneous
value.
Let x  2 .
11  2  2  1
9 2 1
11
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?
?
True. So 2 is the only solution.
Slide 4-6
4.5 Solving an Equation Involving Square
Roots
Graphical Solution The equation in the second step
of the analytic solution has the same solution set as
the original equation. Graph y1  11  x and y 2  1  x
and solve y1 = y2.
The only solution is at x = 2.
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Slide 4-7
4.5 Solving an Equation Involving Cube
Roots
Example Solve 3 x 2  3 x  3 5 .
3
Solution

x  3x 
2
2
3
x  3x
3
5
   5
3
3
3
x  3x  5
2
x  3x  5  0
2
x
3
3  4 (1)(  5 )
2
2 (1)
x
It can be shown that
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3
29
2
the solution set is
-3  2 29 .
Slide 4-8
4.5 Solving an Equation Involving Roots
(Squaring Twice)
Example Solve 2 x  3 
Solution
2x  3 
x 1  1 .
x 1 1
2x  3  1

2x  3

2
 1 
x 1
x 1
Isolate radical.

2
Square both sides.
2x  3  1 2 x 1  x 1
x 1 2 x 1
 x  1   2
2
x 1
Isolate radical.

2
Square both sides.
x  2 x  1  4 ( x  1)
2
x  2x  3  0
2
( x  3 )( x  1)  0
x  1
Write in standard quadratic
form and solve.
or
x3
A check shows that –1 and 3 are solutions of the original equation.
Copyright © 2007 Pearson Education, Inc.
Slide 4-9
4.5 Solving Inequalities Involving Rational
Exponents
Example Solve the inequality 3 x 2  3 x  3 5 .
Solution The associated equation solution in the
previous example was  3 2 29 and  3 2 29 .
Let y  3 x 2  3 x  3 5 and graph y .
Use the x-intercept method
to solve this inequality and
determine the interval where
the graph lies below the xaxis. The solution is the
interval   3 2 29 ,  3 2 29  .
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Slide 4-10
4.5 Application: Solving a Cable
Installation Problem
A company wishes to run a utility cable from point A on the
shore to an installation at point B on the island (see figure).
The island is 6 miles from shore. It costs $400 per mile to run
cable on land and $500 per mile underwater. Assume that the
cable starts at point A and runs along the shoreline, then
angles and runs underwater
to the island. Let x represent
the distance from C at which
the underwater portion of the
cable run begins, and the
distance between A and C be
9 miles.
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Slide 4-11
4.5 Application: Solving a Cable
Installation Problem
(a)
(b)
(c)
(d)
What are the possible values of x in this problem?
Express the cost of laying the cable as a function of x.
Find the total cost if three miles of cable are on land.
Find the point at which the line should begin to angle in
order to minimize the total cost. What is this total cost?
Solution
(a) The value of x must be real where 0  x  9 .
(b) Let k be the length underwater. Using the Pythagorean
theorem,
k
2
 6  x
k 
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2
2
36  x ,
2
k  0.
Slide 4-12
4.5 Application: Solving a Cable
Installation Problem
The cost of running cable is price  miles. If C is the
total cost (in dollars) of laying cable across land and
underwater, then
C ( x )  400 ( 9  x )  500
36  x .
2
(c) If 3 miles of cable are on land, then 3 = 9 – x, giving
x = 6.
C ( 6 )  400 ( 9  6 )  500
36  6
2
 5442 . 64 dollars
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Slide 4-13
4.5 Application: Solving a Cable
Installation Problem
(d) Using the graphing calculator, find the minimum value
of y1 = C(x) on the interval (0,9].
The minimum value of the function occurs when x = 8.
2
So 9 – 8 = 1 mile should be along land, and 36  1  6 . 08
miles underwater. The cost is
C ( 8 )  400 ( 9  8 )  500
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36  8  5400 dollars.
2
Slide 4-14