Chapter 1
Some Additional Info
1
Topics
• Inequalities with absolute values
• Compound inequalities
• Completing the square
• Rational functions
2
Absolute Value
• Remember, |x| is the distance from x to the origin
x can be negative or positive
• In an inequality
|x|<2 means both x<2 and x>-2
it is within two units of the origin
3
More Absolute Value
• So, if we have |x-2| <3, that means
x-2<3 and x-2>-3
equivalently, x<5 and x>-1
-1<x<5
• Check answer: 5-2=3, -2-2=3; the endpoints are correct
if x =0, |0-2|=2<3; the interval is correct
4
And More Absolute Value
• And, if we have |x-2| >3, that means
x-2>3 and x-2<-3, the distance from the origin is >3
or, x>5 or x<-1
• If we want to write this in set notation, we have:
x<-1 ∪ x>5, or we can write (−∞ , -1) ∪ (5, ∞)
• We know the endpoints are right from the previous page
Checking the interval:
x= -2, -2 - 2= - 4 < -3
x = 6, 6 - 2 = 4 > 3
5
Examples
• 3 + |2x + 3| < -5x
• |4-5x| > 2x
• |3-x| > |2-3x|
6
Solutions
3 + |2x + 3| < -5x
(2x+3)<-5x-3 and (2x-3)>5x+3
7x<6, x<6/7 and -3x > 6, x<2
Solution is x<6/7
• |4-5x| > 2x
4x-5>2x or 2x-5 < -2x
2x>5, x>5/2 or 4x<5, x < 5/4,
Solution is x>5/2 or x< 5/4
• |3-x| > |2-3x|
3-x > 2-3x or 3-x>3x-2
2x>-1 or 2x<1
No need to do other pair: 3-x>2-3x and x-3 < 2-3x are already
7
Compound Inequalities
• 3x – 1 < 4x +2 < 8x
• Break into two inequalities
3x-1 < 4x + 2, solution is x > -3
4x + 2 < 8x, 4x>2, x>1/2
• Take the stronger requirement, x>1/2
8
Examples
• 4-x < 3x + 5 < 8x – 1
•
3𝑥+1
2
<
2𝑥+3
3
< 8x
• 3 < -5x + 4 < 13x
9
Solution
• 4-x < 3x + 5 < 8x – 1
4x > 1, x>1/4 and 5x>6, x>6/5
Combine to get x>6/5
•
3𝑥+1
2
<
2𝑥+3
3
< 8x
Eliminate denominator: 9x + 3 < 4x + 6 < 48x
5x<3, x<3/5, and 44x>6, x<3/22
Combine to get x<3/22
• 3 < -5x + 4 < 13x
5x<1 and 4<18x, x<1/5 and x< 9/3
Combine to get x<1/5
10
For answers to compound inequalities
• Can get no solution
• Can get a <x < b
• Can get x< a or x>a
• You can always check your solutions!
11
Completing the Square
1. Isolate the variables
2. Divide by the coefficient of x2
3. Create a Square; watch both sides
4. Take the square root of both sides:
5. Solve for x
3x2 – 6x - 5 = 0
1. 3x2 – 6x = 5
2. x2 – 2x = 5/2
3. (x-1)2 = 5/2 + 1 = 7/2
4. x-1= ±
5. x= 1±
7
7
2
2
12
Examples
• x2 – 5x + 6=0
• 3x2 +11x + 6=0
• 2x2 – 3x - 4=0
13
Solution
• x2 – 5x + 6=0
1. x2 – 5x = - 6
2. x2 – 5x = - 6
3. (x-5/2)2 = - 6 + 5/2 = -7/2
4. x-5/2 =
−7
2
No real solution
14
Solution
• 3x2 +11x - 6=0
1. 3x2 +11x = 6
2. x2 + 11/3 x = 2
3. (x+11/6)2 = 2 + 121/36 = 193/36
4. x+11/6 = ± 193 /6
5. x = 11/6± 193 /6
15
Solution
• 2x2 – 3x - 4=0
1. 2x2 – 3x = 4
2. x2 – 3/2x = 2
3. (x-3/4)2 = 2 + 9/16 = 41
4. x-3/4 = ±
41
16
= ± 41 /4
5. x= 3/4± 41 /4
16
Rational Function
•
3
𝑥
=
1
𝑥+2
17
Solution
•
3
𝑥
=
1
𝑥+2
• Clear fraction by mult by x(x+2)
• 3x + 6 = x or 2x = -6, x = -3
• Not an extraneous solution
18
Rational Function
•
1
𝑥+1
+
1
2𝑥
=
3
𝑥+2
19
Solution
•
1
𝑥+1
+
1
2𝑥
=
3
𝑥+2
• Clear Fractions
2x(x+2) + (x+1)(x+2) = 6x(x+1)
2x2 + 4x + x2 + 3x + 2 = 6x2 + 6x
• Solve for x
3x2 – x – 2 = 0, x = 1, - 2/3
Neither solution is extraneous
20
Rational Function
•
3
𝑧−1
+
𝑧+1
2
=
𝑧
𝑧+2
21
Solution
•
3
𝑧−1
+
𝑧+1
2
=
𝑧
𝑧+2
• To solve, eliminate the fractions. Here multiply by 2(z-1)(z+2)
• 6(z+2) + (z2 -1) = 2z(z-1)
6z + 12 + z2 -1 = 2z2 – 2z
z2 - 8z - 11 =0
z= (8 ± 64 + 44)/2 = 4 ± 108/2
• No Extraneous solutions
• 108 = 9*12 = 3 * 4 * 9, 108/2 =3 3
22