• Share
  • Email
  • Embed
  • Like
  • Save
  • Private Content
Section 1.2 Quadratic Equations
 

Section 1.2 Quadratic Equations

on

  • 639 views

Solving Quadratic Equations

Solving Quadratic Equations

Statistics

Views

Total Views
639
Views on SlideShare
554
Embed Views
85

Actions

Likes
0
Downloads
0
Comments
0

1 Embed 85

http://learn.vccs.edu 85

Accessibility

Upload Details

Uploaded via as Microsoft PowerPoint

Usage Rights

© All Rights Reserved

Report content

Flagged as inappropriate Flag as inappropriate
Flag as inappropriate

Select your reason for flagging this presentation as inappropriate.

Cancel
  • Full Name Full Name Comment goes here.
    Are you sure you want to
    Your message goes here
    Processing…
Post Comment
Edit your comment

    Section 1.2 Quadratic Equations Section 1.2 Quadratic Equations Presentation Transcript

    • Section 1.2Quadratic Equations Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.
    • Quadratic Equations Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.
    • Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.
    • Solve the equation: x − 4x = 0 2 x( x − 4) = 0 x = 0 or x + 4 = 0 x = 0 or x − 4 The solution set is { 0, 4} Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.
    • Solve the equation: x2 = 6 − x x + x−6 = 0 2 ( x + 3)( x − 2) = 0 x = −3 or x = 2 The solution set is { −3, 2} Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.
    • Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.
    • Solve each equation. ( x + 3) 2 (a) x = 7 2 (b) =9 x+3= ± 9 x=± 7 x + 3 = ±3 x = 7 or x = − 7 x + 3 = 3 or x + 3 = −3 x = 0 or x = −6 {The solution set is − 7, 7 } The solution set is { 0, −6} Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.
    • Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.
    • Solve by completing the square: 2x2 + 6x − 5 = 0 2x + 6x = 5 2 3 19 x+ =± 2 4 2x2 6x 5 + = 2 2 2 19 3 x=± − 9 5 9 2 2x + 3x + = + 2 4 2 4 19 − 3 19 − 3 x= or − 2 2 2 3  19x+  = 2 4 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.
    • Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.
    • Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.
    • Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.
    • Use the quadratic formula to find the real solutions if any, of the equation 2x − 4x +1 = 0 2 ax + bx + c = 0 2 b 2 − 4ac = (−4) 2 − 4(2)(1) = 16 − 8 = 8 b − 4ac > 0 so there are two real solutions 2 which can be found using the quadratic formula. Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.
    • Use the quadratic formula to find the real solutions if any, of the equation 2x − 4x +1 = 0 2 −b ± b − 4ac 2 x= 2a −(−4) ± 8 4± 8 2± 2 x= = = 2(2) 4 2 2 + 2 2 − 2    The solution set is  ,   2  2  Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.
    • Use the quadratic formula to find the real solutions if any, of the equation 1 2 x − 6 x + 18 = 0 x 2 − 12 x + 36 = 0 2 ax 2 + bx + c = 0 b 2 − 4ac = (−12) 2 − 4(1)(36) = 144 − 144 = 0 b − 4ac = 0 so there is a repeated solution 2 which can be found using the quadratic formula. Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.
    • Use the quadratic formula to find the real solutions if any, of the equation 1 2 x − 6 x + 18 = 0 x − 12 x + 36 = 0 2 2 ax 2 + bx + c = 0 −b ± b 2 − 4acx= 2a −(−12) ± 0 12 x= = =6 The solution set is { 6} 2(1) 2 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.
    • Use the quadratic formula to find the real solutions if any, of the equation 2x + 3 = 2x 2 Since b − 4ac < 0, 2 2x − 2x + 3 = 0 2 there is no real solution. ax 2 + bx + c = 0 b − 4ac = ( −2 ) − 4(2)(3) 2 2 = 4 − 24 = −20 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.
    • Use the quadratic formula to find the real solutions if any, of the equation 1 2 −1 ± 49 −1 ± 7 6+ − 2 = 0 x= = x x 2(6) 12 6 x2 + x − 2 = 0 −1 + 7 1 = = 12 2 b − 4ac = ( 1) − 4(6)(−2) 2 2 −1 − 7 2 = =− = 1 + 48 = 49 12 3Since b 2 − 4ac > 0, 1 2 The solution set is  , − there are two real solutions. 2 3 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.
    • Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.
    • Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.
    • Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.
    • 9 ( x − 18 ) = 144 ( x − 18) x − 18 = ±4 2 2 = 16 22 centimeters by 22 centimetersx = 18 ± 4 = 22 or 14 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.