Question

Hiya again :) Okay, I don't really get this question. It is my last one. I don't, just want the answer. I'd like the steps, that way I know how to do it, because it will come up again.

http://i57.tinypic.com/14ihjqf.png

Thanks :) I really appreciate it.

Answer 1

What was your attempt?

Answer 2

I had no attempt. I didn't understand it, that's why I'd like for someone to explain it to me :)

Answer 3

There's not much to explain, I'm having a difficult time trying to put it together.

Essentially he put it into a equation that can be factored so he could so for possible solution's. However, one of the solution's is extraneous; 1.

Answer 4

You can see for yourself when you substitute in 1 for "x" and solving.

Answer 5

Do you want to now how they solve the equation or do you want to solve the problem, which the problem ask you to "Show that one of the values does not solve the original radical equation.

Answer 6

-1, sorry. Typo.

Answer 7

Also it wants you to explain how the extra solution arose

Answer 8

Solve the problem. I know, in the steps the person did, there is one wrong. I just need help expaling what went wrong. So with this part "se the logic of equation-solving to explain what went wrong in the process. Explain how the extra solution arose."

Answer 9

*explaining. Sorry.

Answer 10

If you guys don't get it either. I understand :) I'll firgure it out somehow. I just thought help would be ideal.

Answer 11

I think I get it now, the first step, out of the six that was done. Was incorrect. Right?

Answer 12

I will let @tHe_FiZiCx99 answer since he was here first and if I see anything that might be helpful to add I will..

Answer 13

There is nothing wrong with the equation millsemily

Answer 14

Yes there is, there is a uneccasary step. That needs to be indicated.

Answer 15

Afk, soz

The process of "solving" an equation starts with the assumption that there exists a possible solution. But this assumption is sometimes groundless; extraneous. The norm would be to start with an equation, involving a vari., x, and deducing it by a series of steps that the result will be x = z or f, doesn't matter. Assuming that \(\ is \) a root, that root must be either z, f, or whatever. Furthermore, you go on by substituting, when one of the values z, f or etc. in your equation does not satisfy it is called an\(\ extraneous solution\)

Answer 16

\(\ extraneous~solution\) *

Answer 17

The problem asks you " Show one of the values that does not satisfy the original radical equation. quadratic will have 2 or 4 or no roots and sometimes one of those roots or two of the roots will not satisfy the equation.

Answer 18

What I gave you up there can be worded in your answer.

I have to get off. Any question before I do so?

Answer 19

No no, I think I understand it for the most part. Thank you :)

Answer 20

You're welcome ^.~