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Is this true?

  1. Jul 30, 2007 #1
    If [tex]b^{2}-4ac=0 [/tex] and a is not equal to 0, then the graph of [tex]y=ax^{2}+bx +c[/tex] has only one x-intercept.

    I say it is true because, according to the quadratic formula, x would equal [tex]\frac{-b}{2a}[/tex].
  2. jcsd
  3. Jul 30, 2007 #2
    Sure. The quadratic formula tells us the X intercepts are when

    [tex]x=\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/tex]

    So if [tex]b^2 - 4ac = 0[/tex]

    Then we are left with


    Which is a unique solution.
  4. Jul 30, 2007 #3

    i say no.if we let a=0 the x^2 variable vanishes and we are left with a straight line.
    Last edited: Jul 30, 2007
  5. Jul 30, 2007 #4
    Well he specified "and a is not equal to 0" in the original post
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