Download Document

Warm-up
• Given these solutions below: write the
equation of the polynomial:
• 1. {-1, 2, ½)
2 x  3x  3x  2
3
2
Rational Equations
Section 2-6
Objectives
• I can simplify rational
expressions
• I can find Domain
Restrictions
• I can solve rational equations
with one variable
Simplifying Rational Expressions
• Try and reduce numerator over denominator
• You will factor all numerators and
denominators, then
• Reduce or cancel like terms
Domain of Rational Functions
• The domain of any rational function is all
real numbers except where the following
happens:
– No x-value that makes denominator zero
– No x-value that would be a discontinuity (hole)
EXAMPLE 1
Simplify a rational expression
2 – 2x – 15
x
Simplify :
x2 – 9
SOLUTION
x2 – 2x – 15 (x +3)(x –5)
= (x +3)(x –3)
x2 – 9
ANSWER
Factor numerator and denominator.
(x +3)(x –5)
= (x +3)(x –3)
Divide out common factor.
x–5
= x–3
Simplified form
x–5
x–3
Domain Re strictions : x  3, 3
GUIDED PRACTICE
5.
for Examples 1 and 2
x2 – 2x – 3
x2 – x – 6
SOLUTION
x2 – 2x – 3 (x – 3)(x + 1)
x2 – x – 6 = (x – 3)(x + 2)
ANSWER
Factor numerator and denominator.
(x – 3)(x + 1)
= (x – 3)(x + 2)
Divide out common factor.
x+1
= x+2
Simplified form
x+1
x+2
Domain Re strictions : x  2, 3
GUIDED PRACTICE
6.
for Examples 1 and 2
2x2 + 10x
3x2 + 16x + 5
SOLUTION
2x2 + 10x
2x(x + 5)
=
3x2 + 16x + 5 (3x + 1)(x + 5)
=
2x(x + 5)
(3x + 1)(x + 5)
2x
= 3x + 1
ANSWER
2x
3x + 1
Factor numerator and
denominator.
Divide out common factor.
Simplified form
1
Domain Re strictions : x  5, 
3
Adding & Subtracting Rational
Expressions
• MUST have a COMMON DENOMINATOR
• You will factor all denominators, then find the
Common Denominator
• Reduce or cancel like terms
Basic Fraction
2 1

3 2
2
3
1

2
x2
4
x2
x3
6
4+3
3
6
x3
6
7
6
x
2
 2
2
x  5x  6 x  4 x  4
x
( x  2)( x  3)
2

( x  2)( x  2)
(x+2)
(x+2)
x(x+2)
( x  2)( x  3)( x  2)
( x  2)( x  3)( x  2)
(x+3)
2(x+3)
(x+3)
( x  2)( x  3)( x  2)
x 2  2x  2x  6
( x  2)( x  3)( x  2)
x(x+2) - 2(x+3)
x2  6
( x  2)( x  3)( x  2)
x5
x7

2 x  6 4 x  12
x5
2( x  3)
x7

4( x  3)
2
2(x-5)
4( x  3)
2
4( x  3)
1
x-7
1
4( x  3)
2 x  10  x  7
4( x  3)
2(x-5) - 1(x-7)
( x  3)
4( x  3)
1
4
Solving Rational Equations
• Two basic methods
• 1. Set equation equal to ZERO and then get
Common Denominator
• 2. Two ratios equal means you can Cross
Multiply to solve them
Cross Multiplication Method
24
36

x3
x3
36( x  3)  24( x  3)
36x  108  24x  72
12 x  180
x  15
Cross Multiplication Ex2
x2 x4

0
x
x6
( x  2)( x  6)  x( x  4)
x  8 x  12  x  4 x
2
2
 4x  12
x3
x2 x4

x
x6
Set Equation to ZERO
x 1
5x
1


3( x  2) 6 ( x  2)
x 1
3( x  2)
2
x 1
5x
1


0
3( x  2) 6 ( x  2)
2(x+1)
2
6( x  2)
5x

6
(x-2)
5x(x-2)
1

( x  2)
6
(x-2)
6
6( x  2)
6
6( x  2)
2(x+1) – 5x(x-2) - 6
6(x-2)
Next Slide
Problem Continued
2( x  1)  5 x( x  2)  6
0
6( x  2)
2 x  2  5 x 2  10 x  6  0
 5 x 2  12 x  4  0
5 x  12 x  4  0
2
x  12 x  20  0
10
2
( x  )( x  )  0
5
5
2
{2, }
5
MUST CHECK ANSWERS
x = 2 does not work
2
( x  10)( x  2)  0
2
{ }
5
Extraneous Solutions
• Extraneous solutions are those that do not
work when you plug them back into the
original equation.
• Usually they don’t work because they make
the Denominator zero
Homework
• Worksheet 5-1