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Linear Intercepts (lol) / Completing The Square

  1. May 9, 2006 #1
    A couple more i'm having problems with.

    First with linear intercepts.

    Find the y-intercept of a line going through (3, 0) and having a slope of -5


    m = -5

    And from then on, I don't know how to continue with this problem.


    Now, completing the square.

    Solve: Solve: 4x^2 - x - 5 = 0

    4x^2 – x = 5

    x^2 - 1/4x = 5/4

    -1/4 --> 1/16

    x^2 - 1/2x + 1/16 = 5/4 + 1/16

    (x-1/4)^2 = 21/16

    Solve for x now right? Well i think i already messed up in one of my earlier steps. I need help.

    Thanks for any help...
  2. jcsd
  3. May 9, 2006 #2


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    Do you know what b means in y = mx +b?? (hint: that's what you are looking for)

    Check again your work for completing the square.
    Last edited: May 9, 2006
  4. May 9, 2006 #3
    b is the value of the y-intercept.
  5. May 9, 2006 #4


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    The y-intercept is the value y assumes when x = 0. This is when the line cuts through the y-axis. In the equation y = mx + b, what does y equal when x = 0 ?

    You're given a point on the line which is essentially the y-value for a specific x-value. You also know the slope (m). Now put those back in and work out b.
    It's OK up to this step : x^2 - 1/4x = 5/4

    In completing the square, you are basically comparing the left hand side (LHS) to (x-c)^2 = x^2 - 2cx + c^2.

    Now compare the coefficient of the x term in that expansion to your own LHS and find out the value of c. The square of that expression then bears a remarkable resemblance to your original equation, and all you need to do is add constants to make it identical, then you can solve it.
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