More Radical Equations - Resolved

[Xearf_987]

Problem #3

1/ 1 - sqrt(x) = 1 - (sqrt(x)/sqrt(x)-1)

I'm not exactly sure how to start this problem =/

 

[stapel]

I'm not exactly sure how to start this problem

Nor are we, since no instructions were included. Are you perhaps supposed to "solve"...?

Also, your formatting is ambiguous. You have posted this:

. . . . .\(\displaystyle \frac{1}{1}\,-\,\sqrt{x}\,= \,1\,-\,\frac{\sqrt{x}}{\sqrt{x}}\,-\,1\)

...but I have a feeling you probably mean something more like this?

. . . . .\(\displaystyle \frac{1}{1\,-\,\sqrt{x}}\,= \,1\,-\,\frac{\sqrt{x}}{\sqrt{x}\,-\,1}\)

Please reply with clarification, including the instructions.

Thank you.

Eliz.

 

[Xearf_987]

Yes, the second way that you have it written down is what the problem is. And the title is "radical equations"... meaning, you solve for x (obviously). You isolate one radical, you square/cube...etc each member, and repeat this process until you have gotten rid of all the radicals. Then you combine your like terms, and solve for x. This problem has radicals in the denomonaters and I wasn't sure if you're supose to simlify first, or if you find a LCD and multiply it by each member first or what. I hope I've made myself clear this time. If not, then I'll make a third attempt to clarify what the problem is. ><

Thank you for any help you're willing to offer

Also, if you don't mind, can you show show me how to write out my problems the way that you did?

 

[pka]

\(\displaystyle \L
\frac{1}{{1 - \sqrt x }} = 1 - \frac{{\sqrt x }}{{\sqrt x - 1}}\quad \Rightarrow \quad - 1 = \left( {\sqrt x - 1} \right) - \sqrt x \mbox{ if } \quad x \not= 1\)

 

[stapel]

And the title is "radical equations"... meaning, you solve for x (obviously).

Actually, since subject lines frequently have little or nothing to do with the posted question, many tutors have learned not to make any assumptions regarding them.

...can you show show me how to write out my problems the way that you did?

To learn how to format using LaTeX, please follow the links in the "Forum Help" pull-down menu at the very top of the page.

Thank you.

Eliz.

 

[Xearf_987]

Eh? That doesn't look right =/

 

[stapel]

Eh? That doesn't look right

What doesn't "look right"? Please reply with specifics, showing your steps.

Thank you.

Eliz.

 

[Xearf_987]

\(\displaystyle \L
\frac{1}{{1 - \sqrt x }} = 1 - \frac{{\sqrt x }}{{\sqrt x - 1}}\quad \Rightarrow \quad - 1 = \left( {\sqrt x - 1} \right) - \sqrt x \mbox{ if } \quad x \not= 1\)

They isolated the 1 instead of the radical. This, I do not understand.

 

[soroban]

Re: More Radical Equations - Does this make more sense?

Hello, Xearf_987!

\(\displaystyle \text{Problem #3: }\L\:\frac{1}{1\,-\sqrt{x}}\;=\;1\,-\,\frac{\sqrt{x}}{\sqrt{x}\,-\,1}\)

It appears to be an identity . . .


The right side is: \(\displaystyle \L\:1\,-\,\frac{\sqrt{x}}{\sqrt{x}\,-\,1}\;=\;\frac{\sqrt{x}\,-\,1}{\sqrt{x}\,-\,1}\,-\,\frac{\sqrt{x}}{\sqrt{x}\,-\,1} \;= \;\frac{\sqrt{x}\,-\,1\,-\,\sqrt{x}}{\sqrt{x}\,-\,1}\;=\;\frac{-1}{\sqrt{x}\,-\,1}\)


Multiply top and bottom by -\(\displaystyle 1:\;\;\L\frac{-1}{-1}\cdot\frac{-1}{\sqrt{x}\,-\,1} \;=\; \frac{1}{1\,-\,\sqrt{x}}\)

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Or use long division on the left side:

. . . . . . . . . . . \(\displaystyle 1\)
\(\displaystyle \;\;\;1\,-\,\sqrt{x}\;\overline{)\;1\;\;\;\;\;\;\;}\)
. . . . . . . . . . . \(\displaystyle 1\,-\,\sqrt{x}\)
. . . . . . . . . \(\displaystyle \overline{\;\;\;\;\;\sqrt{x}\)

 

[pka]

First you must know that \(\displaystyle \L
1 - \sqrt x = - \left( {\sqrt x - 1} \right)\).

Then because of domain issues it must be true that \(\displaystyle x \not= 1\).

What I showed you is that the equation is true for all x if \(\displaystyle x \not= 1\)

 

[Xearf_987]

Ohh okay, I think I understand now. Thanks a lot!