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Title: PowerPoint Presentation Author: Gerald Wieczorek Last modified by: pete Created Date: 8/29/2005 9:26:49 PM Document presentation format: On-screen Show (4:3) – PowerPoint PPT presentation

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Title: Objectives:


1
  • Objectives
  • Be able to determine if an equation is a rational
    equation.
  • Be able to solve various rational equations and
    exclude any extraneous solutions.

Critical Vocabulary Rational Function,
Extraneous solution
2
  • Rational equation
  • An equation that contains one or more rational
    expressions
  • Solve a rational equation by multiplying both
    sides of the equal sign by the LCD
  • If both sides of the equal sign are rational, you
    can solve be cross multiplying because it will be
    a proportion
  • Least common multiple smallest positive whole
    number exactly divisible by the given numbers

3
I. Rational Functions
A rational equation is an equation that contains
rational expressions (x in the denominator)
Formal Definition A rational Function is a ratio
of two polynomials
written in the form
No, that would be a linear equation.
Looks like you got it.
I think I get it. I bet this is a rational
equation.
What is a Rational Equation?
In Simple terms, its a fraction.
So, would this equation be rational?
So, what is a rational then?
4
II. Solving Rational Equations
b. Solving a Rational By Finding LCM
(Denominator)
What can x not be?
1.
Multiply by LCD
Distribute
5x
6
4
5x 10
This is not an extraneous solution either.
Solution x 2
No.really?
5
II. Solving Rational Equations
a. Solving a Rational Cross Multiplication
(Proportion)
5.
First determine what x cant be
Cross Multiply
(6)(x 5) (x 5)(x 3)
6x30 x2 8x 15
Make it equal 0
0 x2 2x-15
What are talking about? What is an extraneous
solution?
0 (x 5)(x 3)
x -5
x 3
Solution x 3
-5 is an extraneous solution.
Thats where your solution is one of the values
that x cant be.
6
II. Solving Rational Equations
b. Solving a Rational By Finding LCM
(Denominator)
What can x not be?
9.
Multiply by LCD
-x2
5
Distribute
6x
x2 6x5 0
Make it equal 0
(x - 1)(x - 5)0
Both of these solutions look good.
x1 x 5
right
Solution x 1,5
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