- #1

Astro

- 48

- 1

- 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 ##.

(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

Thank you!

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: I**If no, then please explain.**n this example, is**it correct to say that that the two complex values for x are in fact "non-real extraneous solutions"?(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!