Hi All,

I am doing a Precalculus course and this exercise is from schaum's outlines Precalculus textbook.

Please can someone help me understand how this guy came to his conclusion?

Attachment 24124The answerAttachment 24125

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- Jun 21st 2012, 02:00 AMSVDMRadicals and rationalizie
Hi All,

I am doing a Precalculus course and this exercise is from schaum's outlines Precalculus textbook.

Please can someone help me understand how this guy came to his conclusion?

Attachment 24124The answerAttachment 24125 - Jun 21st 2012, 02:19 AMGokuRe: Radicals and rationalizie
$\displaystyle \frac{1}{h}(\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt{x}})$

$\displaystyle =\frac{1}{h}\left(\frac{\sqrt{x}-\sqrt{x+h}}{\sqrt{x(x+h)}}\right)$

$\displaystyle =\frac{1}{h}\left(\frac{\sqrt{x}-\sqrt{x+h}}{\sqrt{x(x+h)}}\right)\left( \frac{\sqrt{x}+\sqrt{x+h}}{\sqrt{x}+\sqrt{x+h}} \right)$

$\displaystyle =\frac{1}{h}\left(\frac{x -x-h}{\sqrt{x(x+h)}}\right)\left( \frac{1}{\sqrt{x}+\sqrt{x+h}} \right)$

Now continue from there....(Evilgrin) - Jun 21st 2012, 02:41 AMbiffboyRe: Radicals and rationalizie
- Jun 21st 2012, 02:56 AMSVDMRe: Radicals and rationalizie
Hallo ou maat!

My head is spinning!

Thanks very much as I analyze this for the next 3 hours....

I never seen this done before! Not in the textbook anyway.

Stan - Jun 21st 2012, 03:51 AMGokuRe: Radicals and rationalizie
Remember that $\displaystyle a^2 - b^2 = (a-b)(a+b)$...

- Jun 21st 2012, 12:03 PMSVDMRe: Radicals and rationalizie
Hi Guys thanks for the help!

I have another one that I can not solve. I don't find much help in text book and it does not help trying my own dodgy thinks so I hope to learn something from you all.

This time the stuffed me up with this...

Attachment 24131

? I tried to solve this but the answer is so wrong... - Jun 21st 2012, 12:16 PMPlatoRe: Radicals and rationalizie
- Jun 21st 2012, 12:19 PMSVDMRe: Radicals and rationalizie
Here ia another problem that keep me from sleeping.

Attachment 24132

what I do next brings me to the answer but i think it is dodgy and Ii would like to see how it is really doubt so Ii can be sure it is right.

Thanks again. - Jun 21st 2012, 02:04 PMReckonerRe: Radicals and rationalizie
I don't know what's going on in your denominator...

$\displaystyle \frac{\sqrt{x+1} - \sqrt{a+1}}{x-a}$

$\displaystyle =\frac{\sqrt{x+1} - \sqrt{a+1}}{x-a}\cdot\frac{\sqrt{x+1} + \sqrt{a+1}}{\sqrt{x+1} + \sqrt{a+1}}$

$\displaystyle =\frac{(x+1) - (a+1)}{(x - a)\left(\sqrt{x+1} + \sqrt{a+1}\right)}$

$\displaystyle =\frac{x - a}{(x - a)\left(\sqrt{x+1} + \sqrt{a+1}\right)}$

$\displaystyle =\frac1{\sqrt{x+1} + \sqrt{a+1}}$ - Jun 22nd 2012, 03:31 PMSVDMRe: Radicals and rationalizie
Hi again,

I am looking through the text book to find how you came to the conclusion to use a square of a difference in this case. the radical is cubed so I would thought to use differences of two cubes?

Or am I confused with the radicals somehow? If I disregard manipulating the numerator and just assume the difference of two cubes will remove the radicals my answer is correct. But I am missing something seriously with my understanding of polynomial manipulation because when I actually do the calculation. I get the result below which shows that something is not correct. (Crying)

Attachment 24136 - Jun 22nd 2012, 06:53 PMskeeterRe: Radicals and rationalizie
to rationalize the numerator of the fraction ...

$\displaystyle \frac{\sqrt[3]{x} - \sqrt[3]{a}}{x-a}$

note that the denominator can be factored as the difference of two cubes ...

$\displaystyle x - a = (\sqrt[3]{x} - \sqrt[3]{a}) \left[(\sqrt[3]{x})^2 + \sqrt[3]{x} \cdot \sqrt[3]{a} + (\sqrt[3]{a})^2 \right]$

can you finish? - Jun 22nd 2012, 08:39 PMSVDMRe: Radicals and rationalizie
- Jun 23rd 2012, 02:26 PMSVDMRe: Radicals and rationalizie
Please can you explain exactly why you came to this conlusion?

I am trying to understand what to look for to solve the problem myself but getting the answer still does not explain the process it takes to reach the conclusion.

Thanks again. - Jun 23rd 2012, 02:39 PMPlatoRe: Radicals and rationalizie