Square root function: x intercept paradox?

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Homework Help Overview

The discussion revolves around finding the x-intercept of the square root function defined by the equation y = √(x + 4) + 1. The original poster attempts to solve for the x-intercept by setting y to 0, leading to confusion regarding the validity of the resulting solution.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the implications of setting the square root expression to a negative value and the resulting solutions that may not be real. There is also inquiry into whether the roots can be classified as imaginary.

Discussion Status

The discussion is ongoing, with participants providing insights into the nature of square root functions and the consequences of squaring both sides of an equation. Some guidance has been offered regarding the interpretation of x-intercepts in the context of the function's graph.

Contextual Notes

There is a focus on the distinction between real and imaginary solutions, as well as the impact of transformations on the graph of the function. The original poster's confusion stems from the relationship between the algebraic manipulation and the graphical representation of the function.

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.)
 
Last edited:

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