Algebra 4'6 - PowerPoint PPT Presentation

1 / 31
About This Presentation
Title:

Algebra 4'6

Description:

x2 2x - 1224 = 0 ( x 36) ( x 34) =0. x = -36, 34. So -36 and -34 and 34 and 36 ... results in a disjunction / or statement (x n or x n) ... – PowerPoint PPT presentation

Number of Views:20
Avg rating:3.0/5.0
Slides: 32
Provided by: Sim4127
Category:

less

Transcript and Presenter's Notes

Title: Algebra 4'6


1
Algebra 4.6
  • Solving Quadratic Inequalities

2
Homework Review
3
Homework Review
  • 4. x (x 2 ) 1224
  • x2 2x - 1224 0
  • ( x 36) ( x 34) 0
  • x -36, 34
  • So -36 and -34 and 34 and 36

4
Homework Review
  • 6. x (x - 9 ) 736
  • x2 - 9x - 736 0
  • ( x 32) ( x 23) 0
  • x 32, -23
  • So 32 and 23 and -23 and -32

5
Homework Review
  • 8. (702x) (502x ) 4524
  • 3500240x 4x2 4524
  • 4x2 240x - 1024 0
  • (x 64) ( x - 4 )0
  • X -64, 4
  • So X has to be 4

x
50
x
70
6
Homework Review
  • 10. 122 202 x2
  • 144 400 x2
  • 544 x2
  • x /- v544
  • So X is ?23.3 ft

x
20
12
7
Homework Review
x by x
  • 12. (18 2x)(13-2x) 176
  • 234 - 62x 4x2 176
  • 4x2 - 62x 58 0
  • 2x2 - 31x 29 0
  • ( x 1) (2x 29) 0
  • X 1 0r 14.5
  • So X 1 in

13
18
8
Homework Review
  • 14. (14 2x)(122x) 288
  • 168 52x 4x2 288
  • 4x2 52x - 120 0
  • x2 13x - 30 0
  • (x 15) (x 2) 0
  • X -15 0r 2
  • So X 2 in

12
x
14
x
9
Solving Quadratic Inequalities
  • Compound Inequality Theorem (lesson 2.8)

10
Solving Quadratic Inequalities
  • Compound Inequality Theorem (lesson 2.8)
  • gt results in a disjunction / or statement (x lt n
    or x gt n)

11
Solving Quadratic Inequalities
  • Compound Inequality Theorem (lesson 2.8)
  • gt results in a disjunction / or statement (x lt n
    or x gt n)
  • lt results in a conjunction / and statement (n lt x
    lt x)

12
Solving Quadratic Inequalities
  • Example
  • 15x2 26x 80 gt 0
  • (The book tells you to solve this by completing
    the square this will allow you to easily rewrite
    as an absolute value equation but the formula
    will work as well.)

13
Solving Quadratic Inequalities
  • Example
  • 15x2 26x 80 gt 0
  • to complete the square
  • 15x2 26x gt 80

14
Solving Quadratic Inequalities
  • Example
  • 15x2 26x 80 gt 0
  • to complete the square
  • 15x2 26x gt 80
  • x2 (26/15)x gt (80/15)

15
Solving Quadratic Inequalities
  • Example
  • 15x2 26x 80 gt 0
  • to complete the square
  • 15x2 26x gt 80
  • x2 (26/15)x gt (80/15)
  • x2 (26/15)x (676/30) gt (80/15) (676/900)

16
Solving Quadratic Inequalities
  • Example
  • 15x2 26x 80 gt 0
  • to complete the square
  • 15x2 26x gt 80
  • x2 (26/15)x gt (80/15)
  • x2 (26/15)x (676/30) gt (80/15) (676/900)
  • (x (26/30))2 gt (5476/900)

17
Solving Quadratic Inequalities
  • Example
  • 15x2 26x 80 gt 0
  • to complete the square
  • 15x2 26x gt 80
  • x2 (26/15)x gt (80/15)
  • x2 (26/15)x (676/30) gt (80/15) (676/900)
  • (x (26/30))2 gt (5476/900)
  • x (26/30) gt /- 74/30

18
Solving Quadratic Inequalities
  • Example
  • 15x2 26x 80 gt 0
  • to complete the square
  • 15x2 26x gt 80
  • x2 (26/15)x gt (80/15)
  • x2 (26/15)x (676/30) gt (80/15) (676/900)
  • (x (26/30))2 gt (5476/900)
  • x (26/30) gt /- 74/30
  • X gt /- (74/30) (26/30)

19
Solving Quadratic Inequalities
  • Example
  • 15x2 26x 80 gt 0
  • to complete the square
  • 15x2 26x gt 80
  • x2 (26/15)x gt (80/15)
  • x2 (26/15)x (676/30) gt (80/15) (676/900)
  • (x (26/30))2 gt (5476/900)
  • x (26/30) gt /- 74/30
  • X gt /- (74/30) (26/30)
  • Xgt 48/30 or x gt -100/30

20
Solving Quadratic Inequalities
  • Example
  • 15x2 26x 80 gt 0
  • to complete the square
  • 15x2 26x gt 80
  • x2 (26/15)x gt (80/15)
  • x2 (26/15)x (676/30) gt (80/15) (676/900)
  • (x (26/30))2 gt (5476/900)
  • x (26/30) gt /- 74/30
  • X gt /- (74/30) (26/30)
  • Xgt 48/30 or x gt -100/30
  • X lt -10/3 or x gt 8/5

21
Solving Quadratic Inequalities
  • Example
  • 15x2 26x 80 gt 0
  • The final solution looks like this)
  • X lt -10/3 or x gt 8/5

22
Solving Quadratic Inequalities
  • Example
  • x2 4x 21 lt 0
  • (Solve this and get your two solutions)
  • X -7 and 3
  • (Since this is a conjunction the final solution
    looks like this)
  • -7 lt x lt 3

23
Additional Practice
  • Solve each inequality
  • X2 x 20 gt 0
  • X2 10x 16 lt 0
  • 5x2 10 gt 27x
  • 9x2 31x 12 lt 0

24
Additional Practice
  • Solve each inequality
  • X2 x 20 gt 0 x gt 5 or x lt -4
  • X2 10x 16 lt 0 2 lt x lt 8
  • 5x2 10 gt 27x x gt 5 or x lt 2/5
  • 9x2 31x 12 lt 0 -3 lt x lt -4/9

25
Expressing as absolute value inequalities
  • You must use the completing the square method
  • Look at Example 1

26
Expressing as absolute value inequalities
  • 15x2 26x 80 gt 0
  • to complete the square
  • 15x2 26x gt 80
  • x2 (26/15)x gt (80/15)
  • x2 (26/15)x (676/30) gt (80/15) (676/900)
  • (x (26/30))2 gt (5476/900)

27
Expressing as absolute value inequalities
  • 15x2 26x 80 gt 0
  • When we completed the square
  • (x 26/30)2 gt 5476/900
  • Then rewrite as x26/30 gt 74/30
  • Or simplified x13/15 gt 37/15

28
Practice Expressing as absolute value inequalities
  • Start to solve by completing the square and then
    rewrite as an absolute value inequality
  • X2 6x - 40 lt 0

29
Practice Expressing as absolute value inequalities
  • Solve by completing the square and then rewrite
    as an absolute value inequality
  • X2 6x - 40 lt 0
  • X2 6x 9 lt 40 9

30
Practice Expressing as absolute value inequalities
  • Solve by completing the square and then rewrite
    as an absolute value inequality
  • X2 6x - 40 lt 0
  • X2 6x 9 lt 40 9
  • (x 3)2 lt 49

31
Practice Expressing as absolute value inequalities
  • Solve by completing the square and then rewrite
    as an absolute value inequality
  • X2 6x - 40 lt 0
  • X2 6x 9 lt 40 9
  • (x 3)2 lt 49
  • x - 3 lt 7
Write a Comment
User Comments (0)
About PowerShow.com