Solving radical equations: Do non-real extraneous solutions exist?

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In solving the radical equation x + 4 = √(x + 10), the discussion centers on whether the complex solutions derived from squaring the equation can be classified as "non-real extraneous solutions." It is established that the complex solutions x = 1/2(-9 ± √23i) do not satisfy the original equation, thus they cannot be considered extraneous solutions. The term "extraneous solution" applies only to valid solutions of a derived equation that do not solve the original problem. The conversation also highlights the importance of correctly interpreting the square root and absolute value in the context of complex numbers. Overall, the classification of complex solutions as extraneous depends on their relationship to the original equation.
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Homework Statement
Personal study question:

Is it correct to say that in this example, the two complex values for x are in fact "non-real extraneous solutions"?
Relevant Equations
Original equation: x+4 = √(x+10) .
(See work below for further equations.)
Assume you have a radial equation (eg. x+4 = √(x+10) ) that you want to solve for "x".

To solve:
\begin{align} (x+4 & = \sqrt[2] {x+10})^2 \nonumber \\
(x+4)^2 &= |x+10| \nonumber \end{align}
For my question, we are only going to consider the case where x+10 < 0:
\begin{align} x^2+8x+16 &= -(x+10) \nonumber \\
x^2+8x+16 &= -x-10 \nonumber \\
0 &= x^2+9x+26 \nonumber \end{align}

Using the quadratic formula, we get:

\begin{align} x &= \frac {-9 \pm \sqrt{9^2 -4(1)(26)}} {2(1)} \nonumber \\
x &= \frac {-9 \pm \sqrt{-23}} {2} \nonumber \\
x &= \frac{1}{2}(-9 \pm \sqrt{23}i) \nonumber \end{align}

After checking if LS = RS for the original equation for each complex answer, it seems that neither is a solution because ## LS \neq RS ##.

Therefore, what I'd like to know is: In this example, is it correct to say that that the two complex values for x are in fact "non-real extraneous solutions"? If no, then please explain.

(I would also appreciate if anyone has an links to articles about extraneous non-real solutions to radical equations. Most of the articles online that I can easily find basically only talk about extraneous solutions in the context of "real number" answers. I'm not really sure if it makes any mathematical sense to about extraneous complex solutions--I don't know enough about the subject. To be clear, I already know that "extraneous solutions" are invalid solutions to the starting equation; I just don't know if the term "extraneous solutions" can be also applied to non-real solutions that happen to be invalid solutions. In other words, I'm having trouble visualizing what a complex extraneous solution would look like graphically--which makes me wonder if such a thing even exists. )

Thank you!
 
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Whether complex solutions are "extraneous" or not depends on the problem that this equation is being applied to. Complex solutions certainly can have a real-world meaning in the right situation. The entire complex plane can have a physical meaning, such as phase and gain margins in a feedback circuit.
CORRECTION: I missed the point that the complex values were not really solutions to the original problem. They were probably introduced when both sides were squared.
 
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Astro said:
In this example, is it correct to say that that the two complex values for x are in fact "non-real extraneous solutions"?
No. In order for ## x = w ## to be an extraneous solution, it needs to be a solution and you have shown that neither value of ## x = \frac{1}{2}(-9 \pm \sqrt{23}i) ## is a solution.
Astro said:
In other words, I'm having trouble visualizing what a complex extraneous solution would look like graphically
An example would be the solution ## \frac{\sqrt 3 i - 1}2 \mathrm{m} ## to "how long are the sides of a cube with volume ## 1 \mathrm{m}^3##?"
 
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Astro said:
To be clear, I already know that "extraneous solutions" are invalid solutions to the starting equation
But this is not correct; an extraneous solution is a solution* to an equation that is not a solution to the underlying problem.

* There is no such thing as an 'invalid solution'. Either a candidate value is a solution or it is not.
 
pbuk said:
But this is not correct; an extraneous solution is a solution* to an equation that is not a solution to the underlying problem.
Quite so, but that includes solutions to a derived equation that are not solutions to the original equation. See e.g. https://en.m.wikipedia.org/wiki/Extraneous_and_missing_solutions.
When @Astro replaced ##\sqrt{x+10}^2## with ##|x+10|## it admitted two extraneous solutions. I see no objection to describing them as "non-real extraneous solutions".

That said, there are a couple of issues with the algebra. If complex solutions are going to be allowed then ##\sqrt{x+10}^2## is not the same as |x+10|, and neither is ##\sqrt{(x+10)^2}##.
Moreover, there was no ambiguity in the first place. ##\sqrt{x+10}^2=x+10##. No ##\pm## arises. On the other hand, ##\sqrt{(x+10)^2}=\pm(x+10)##.
 
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