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Square root function: x intercept paradox?

  1. Jul 29, 2015 #1
    Here's my equation y= sqrt (x+4)+1

    I want to find the x intercept. Ok so i replace y with 0 and solve for x
    0= √(x+4)+1
    -1=√(×+4)
    (-1)2=×+4
    1=×+4
    -3=×

    So it looks like the x intercept is (-3,0)

    But then when i go back and plug in -3 in place of x...
    y=√(-3+4)+1
    y=1+1
    y=2

    I now have a point (-3,2)!

    I checked the graph and it accepts the point (-3,2). So whats wrong with my first work putting in 0 for y and giving me (-3,0)? Please help me understand
     
  2. jcsd
  3. Jul 29, 2015 #2

    mfb

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    Here is the problem. The square root cannot be negative so the first line does not have solutions.
    By squaring you create additional solutions to the equation which are not real.
     
  4. Jul 29, 2015 #3
    Thx for the response!

    So then, is it classified as an imaginary root?
     
  5. Jul 29, 2015 #4

    mfb

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    If you look for x intercepts, you probably look for real values. There are imaginary values of x where the function (seen as complex function) is zero, yes.
     
  6. Aug 1, 2015 #5
    The thing with square root functions is whenever you attempt to solve for x, you end up turning them into conical parabolas. This means you can get x-intercepts for the conical parabola which don't actually exist for the square root function.
     
  7. Aug 8, 2015 #6

    symbolipoint

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    You have the upper branch of a square root function, and it has been shifted UPWARD by 1 unit, and therefore what once was an x-axis intercept has been stopped; as well as any intersection with the x-axis.

    (I made a drawing and saved, using "whiteboard" but the file is missing and is therefore not available to upload.)
     
    Last edited: Aug 8, 2015
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