1.Create an equation that results in at least one extraneous solution. Work through your equation, justify each step, and explain how the solution is extraneous.

Question

anonymous · Answer

I don't know what extraneous means, or what a extraneous solution is. Can some one help me with this?

anonymous · Answer

@hartnn  @satellite73  @Hero  i'm sorry to bother you guys but can any of you help?

hero · Answer

There are links all over the web that explain what an extraneous solution is. Have you tried doing a google search?

anonymous · Answer

I did some research on it and it seems that a extraneous solution is a solution that can not be possible? i'm a little lost, i'm sorry :(

hero · Answer

See if you can understand this: http://hotmath.com/hotmath_help/topics/extraneous-solutions.html

anonymous · Answer

that's actually the page i'm on now, but I can't really understand what they mean by a root of a transformed equation that is not a root of the original equation because it was excluded from the domain of the original equation. I'm sorry.

hero · Answer

Try plugging the extraneous solution back in to the original equation to check it.

hero · Answer

There are certain rules for determining the domain of a function. If you learned how to do this, you could figure out what the domain of the function is before attempting to solve the problem.

hero · Answer

Then after you solve the problem, if your result is out of the domain, then you know that it is not a solution.

hero · Answer

But you can also use the check method to figure out if your result is extraneous.

anonymous · Answer

You know what, I just checked my school and it seems that I skipped a lesson. I'm going to look at that lesson to see if it teaches me about it, so I will come back after I read it. I'm sorry again

anonymous · Answer

hello, i'm back and I was wondering if you can check this problem I made up to see if it is a correct extraneous solution?

hero · Answer

Okay, post it...

anonymous · Answer

ok so I figured out that extraneous solution I like solving the problem but when you fill it in, it doesn't equal to the right number
so
$$\sqrt{x - 6}+8 = 2$$

anonymous · Answer

$$\sqrt{x-6}=-6$$

anonymous · Answer

$$\left(\begin{matrix}\sqrt{x-6} \ \end{matrix}\right)^2 = (-6)^2$$

anonymous · Answer

$$x-6 = 36$$

anonymous · Answer

$$x = 42$$

anonymous · Answer

then when you fill in 42 for x it is
$$\sqrt{42-6}+8 = 2$$

anonymous · Answer

$$\sqrt{44}\neq 2$$

hero · Answer

Actually you should have gotten

$\sqrt{36} = 2$
     $6 \ne 2$

hero · Answer

Or rather...
$6 \ne -6$

anonymous · Answer

but isn't it when x -6 = 36 that when you add 6 on both sides you get x = 44

anonymous · Answer

not 44 I mean 42

hero · Answer

You didn't check correctly...
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