Square root function: x intercept paradox?

AI Thread Summary
The discussion revolves around finding the x-intercept of the function y = √(x + 4) + 1. When substituting y with 0, the calculation incorrectly suggests an x-intercept at (-3, 0), which is not valid since the square root cannot yield a negative value. The correct point found by substituting x = -3 back into the function is (-3, 2), indicating that the function does not intersect the x-axis. The confusion arises from squaring both sides of the equation, which introduces extraneous solutions that do not apply to the original square root function. Ultimately, the function is shifted upward, preventing any real x-intercepts.
Dapperdub
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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 what's wrong with my first work putting in 0 for y and giving me (-3,0)? Please help me understand
 
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Dapperdub said:
-1=√(×+4)
(-1)2=×+4
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.
 
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mfb said:
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.

Thx for the response!

So then, is it classified as an imaginary root?
 
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.
 
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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.
 
Dapperdub said:
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 what's wrong with my first work putting in 0 for y and giving me (-3,0)? Please help me understand
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.)
 
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