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-   -   Solve by completing square (https://www.askmehelpdesk.com/showthread.php?t=2621)

  • Jun 13, 2004, 05:54 PM
    figureen
    solve by completing square
    2x^2-3x=1
  • Aug 12, 2004, 09:45 PM
    rohan
    Re: solve by completing square
    Quote:

    2x^2-3x=1
    2x²-3x=1
    so 2x²-3x-1=0 ------ let A

    let ax²+bx+c=0
    hence x=(-b±?(b²-4ac))/2a ------ let B

    here from A & B

    a=2, b=-3, c=-1

    so x=[-(-3)±?((-3)²-4.2.-1)]/2.2

    hence x=[3±?(9+8)]/4

    x=[3±?17]/4

    hence x1=[3+?17]/4 or x2=[3-?17]/4

    x1=[3+4.123]/4 or x2=[3-4.123]/4


  • Aug 12, 2004, 09:52 PM
    rohan
    Re: solve by completing square
    Quote:


    2x²-3x=1
    so 2x²-3x-1=0 ------ let A

    let ax²+bx+c=0
    hence x=(-b±?(b²-4ac))/2a  ------ let B

    here from A & B

    a=2, b=-3, c=-1

    so x=[-(-3)±?((-3)²-4.2.-1)]/2.2

    hence x=[3±?(9+8)]/4

             x=[3±?17]/4

    hence x1=[3+?17]/4  or x2=[3-?17]/4

             x1=[3+4.123]/4  or x2=[3-4.123]/4
       
    x1=1.78 or x2=-0.28
    javascript:rolleyes()


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