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Thread: 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 $\displaystyle ax^2 + bx + c = 0$ then its solutions are

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

    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 $\displaystyle \Delta = b^2 - 4ac$.

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

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

    If $\displaystyle \Delta < 0$ 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 $\displaystyle x$, it was kind of assumed that the other variable is $\displaystyle y$. This would mean that the graph that the question talks about is the graph whose equation is

    $\displaystyle y = x^2+12x+36$

    Now the $\displaystyle x$-axis is horizontal, and the $\displaystyle y$-axis is vertical. So the numbers on the $\displaystyle x$-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 $\displaystyle y$-axis tell you how far up or down you go from the origin: positive up, negative down.

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


    $\displaystyle x^2+12x+36 = 0$

    Is this OK now?

    Grandad
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