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

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Thus the first equation admits extraneous solutions.In summary, the original equation has two pairs of solutions, both with extraneous solutions. The first pair is (a) the trivial solution x=-10, for which ##-10+10=0##, and (b) the solution x=6, for which ##6+10=4^2=16##. The second pair is (a) the solution x=(-9+√23i)/2, for which ##(-9+√23i+10)^2=26##, and (b) the solution x=(-9-√23i)/2, for which ##(-9-√23i+10)^2=26
  • #1
Astro
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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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  • #2
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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  • #3
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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  • #4
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.
 
  • #5
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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Related to Solving radical equations: Do non-real extraneous solutions exist?

1. What is a radical equation?

A radical equation is an equation that contains a variable inside a radical symbol, such as a square root or cube root.

2. How do you solve a radical equation?

To solve a radical equation, you must isolate the radical term on one side of the equation and then square both sides to eliminate the radical symbol. This process may need to be repeated multiple times if there are multiple radical terms.

3. What are extraneous solutions?

Extraneous solutions are solutions that appear to satisfy the original equation, but when plugged back into the equation, do not actually make the equation true. These solutions often occur when squaring both sides of a radical equation.

4. Can non-real extraneous solutions exist?

Yes, non-real extraneous solutions can exist in radical equations. This means that the solutions involve imaginary numbers, such as the square root of a negative number.

5. How can you check for extraneous solutions?

To check for extraneous solutions, you can plug the solutions back into the original equation and see if they make the equation true. If a solution does not make the equation true, then it is an extraneous solution.

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