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Express g(x) as g(x) = -2x2+ 0x + 1

  1. Oct 24, 2004 #1
    Ok i have the following question:

    Given the functions:

    F(x) = 2x^2 + 3x - 2
    G(x) = 1 - 2x^2

    Find:

    a) the zeros of f(x), g(x)

    Now ive used the following formula
    Code (Text):

    [U]-b ±√b²-4ac²[/U]
            2a
    and worked out the zeros of f(x) fine, but I'm confused as to how to accomplish the same for g(x) when g(x) only has 2 elements. Any help?
     
  2. jcsd
  3. Oct 24, 2004 #2
    dont u just set g(x) = 0? if so you get [tex]x = \frac{\sqrt{2}}{2}[/tex]
     
  4. Oct 24, 2004 #3

    arildno

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    Why do think g has "only two elements"?
     
  5. Oct 24, 2004 #4

    Math Is Hard

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    I hope I remember this correctly, but I believe you can express g(x) as
    g(x) = -2x2+ 0x + 1

    also, the c should not be squared in your quadratic formula.
     
  6. Oct 24, 2004 #5
    yup thanks that was a typo
     
  7. Oct 24, 2004 #6
    for g(x) :

    x= +/- root2/2
     
  8. Oct 24, 2004 #7

    dav2008

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    You don't need the quadratic formula for g(x)

    If you want to find the zeros:

    g(x)=1 - 2x2

    0=1 - 2x2

    2x2=1
    x2=1/2

    Then, like faraz said [tex]x = \frac{\sqrt{2}}{2}[/tex]

    but also

    [tex]x = -\frac{\sqrt{2}}{2}[/tex]
     
  9. Oct 24, 2004 #8
    oh i see. ok. thanks for your help guys. much appreciated.
     
  10. Oct 24, 2004 #9
    how do you plot [tex]x = \frac{\sqrt{2}}{2}[/tex] on a graph though?
     
  11. Oct 24, 2004 #10

    dav2008

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    Why would you want to?

    I guess if you really wanted to plot it an a x-y coordinate plane it would just be a vertical line at [tex]x = \frac{\sqrt{2}}{2}[/tex]
     
  12. Oct 24, 2004 #11
    you don't need the negative in front of the fraction, because the result of the radical is automatically assumed to be plus and minus.

    [tex]\sqrt{2} = \pm1.414...[/tex]
     
  13. Oct 24, 2004 #12

    dav2008

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    I believe the square root function (being a function and all) only outputs the positive root.
     
  14. Oct 24, 2004 #13
    the function we are originally dealing with is not the square root function, it's g(x)=1 - 2x^2. Therefore, you have to include both + and - values of g(0). If you plot the graph, it is a parabola, so will have one, two, or no roots. In this case, it has two, one positive and one negative. Hence the +- of your square root.
     
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