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Math Help - Crossing the x-axis

  1. #1
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    Question Crossing the x-axis

    I'm not really sure how to start this.

    How many times does this cross the x-axis:

    a) x^2 + 12x + 36 = 0

    Does the 0 make a difference, or would I just see what multiplys to get 36 and adds to get 12 and then factor it that way? This is important and I'm totally lost, I can't remember and I can't find any notes whatsoever on how to do equations that = 0, or equations that ask how many times they cross the x-axis.

    Any help would be appreciated!
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  2. #2
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    Quote Originally Posted by ~NeonFire372~ View Post
    I'm not really sure how to start this.

    How many times does this cross the x-axis:

    a) x^2 + 12x + 36 = 0

    Does the 0 make a difference, or would I just see what multiplys to get 36 and adds to get 12 and then factor it that way? This is important and I'm totally lost, I can't remember and I can't find any notes whatsoever on how to do equations that = 0, or equations that ask how many times they cross the x-axis.

    Any help would be appreciated!
    The x-axis occurs at y = 0, so the x-axis intercepts are found by letting y = 0. So yes the = 0 is important, no quadratic equation can be solved without it.

    If you have a quadratic of the form ax^2 + bx + c = 0 then its solutions are

    x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.

    I'll leave you to do the research why this is so - hint, you complete the square.


    Can you see that the solution depends on stuff under a square root? This is called the discriminant, and is denoted by \Delta = b^2 - 4ac.

    If \Delta > 0 there are exactly two solutions. (Why?)

    If \Delta = 0 there is exactly one solution. (Why?)

    If \Delta < 0 there are not any solutions. (Why?)


    Can you see what a, b and c are in this equation? Find the discriminant and tell me what you can determine about the solutions.
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  3. #3
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    I don't really get what you're saying...what would be the first thing I'd do here? There's no y in the equation that I can make equal 0...

    I looked around in my folder a bit more and found a sheet of paper with something similar to that there, so I'm gonna try it that way and see if it works:

    x^2 + 12x + 36 = 0
    (x+6)(x+6) since 6+6=12 and 6*6=36 = 0

    (x+6)=0 (x+6)=0
    x=-6 x=-6

    There's only one solution so it crossed the x-axis once.

    Is that right?
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  4. #4
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    yes correct......
    if the descriminant D= b^2-4ac of a quadratic equation

    1) is > 0 then it has 2 distinct real roots and it crosses the x axis at 2 distinct points

    2) and if it is equal to zero (as in this question) then it has only one real root and it TOUCHES the x axis at a single point

    3)and if D<0 then it has no real roots and the curve doesnt crosses/touches the x axis
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  5. #5
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    Hello ~NeonFire372~
    Quote Originally Posted by ~NeonFire372~ View Post
    There's no y in the equation that I can make equal 0...
    It only makes sense to talk about a graph when you have an equation involving two variables. Since the equation you have involves a variable called x, it was kind of assumed that the other variable is y. This would mean that the graph that the question talks about is the graph whose equation is

    y = x^2+12x+36

    Now the x-axis is horizontal, and the y-axis is vertical. So the numbers on the x-axis tell you how far you move from the origin (0, 0) to the left or right, positive to the right, negative to the left. And the numbers on the y-axis tell you how far up or down you go from the origin: positive up, negative down.

    The y-values, then, tell you how far you are above or below the x-axis. So, if you want to know where any graph crosses the x-axis, you find the values of x (if there are any) that make the value of y zero. So in this case you need to find the value(s) of x for which


    x^2+12x+36 = 0

    Is this OK now?

    Grandad
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