Radicals and rationalizie

**SVDM** · June 21st 2012, 02:00 AM

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?

The answer

**Goku** · June 21st 2012, 02:19 AM

$\frac{1}{h}(\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt{x}})$

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

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

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

Now continue from there....

**biffboy** · June 21st 2012, 02:41 AM

Originally Posted by SVDM

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?

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The answer

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Multiply numerator and denominator by root(x+h)root(x) Then multiply numerator and denominator by root(x)+root(x+h)

**SVDM** · June 21st 2012, 02:56 AM

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

**Goku** · June 21st 2012, 03:51 AM

Remember that $a^2 - b^2 = (a-b)(a+b)$ ...

**SVDM** · June 21st 2012, 12:03 PM

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...

? I tried to solve this but the answer is so wrong...

**Plato** · June 21st 2012, 12:16 PM

Originally Posted by SVDM

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...

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Multiply by numerator and denominator by $\left( {\sqrt[3]{{{x^2}}} - \sqrt[3]{x}\sqrt[3]{a} + \sqrt[3]{{{a^2}}}} \right)$

**SVDM** · June 21st 2012, 12:19 PM

Here ia another problem that keep me from sleeping.

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.

**Reckoner** · June 21st 2012, 02:04 PM

Originally Posted by SVDM

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.

I don't know what's going on in your denominator...

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

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

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

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

$=\frac1{\sqrt{x+1} + \sqrt{a+1}}$

**SVDM** · June 22nd 2012, 03:31 PM

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.

Radicals and rationalizie-radical-polynomial-issue.jpg

**skeeter** · June 22nd 2012, 06:53 PM

to rationalize the numerator of the fraction ...

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

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

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

can you finish?

**SVDM** · June 22nd 2012, 08:39 PM

Hi thanks,

So I did misunderstand Pluto above;

Originally Posted by Plato

Multiply by numerator and denominator by $\left( {\sqrt[3]{{{x^2}}} - \sqrt[3]{x}\sqrt[3]{a} + \sqrt[3]{{{a^2}}}} \right)$

Now I see what was done! I feel so stupid! How could I have known? Looks like I have to study the factoring rules some more!

Thank you very much.

**SVDM** · June 23rd 2012, 02:26 PM

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.

**Plato** · June 23rd 2012, 02:39 PM

Originally Posted by SVDM

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.

You must know that $x^3\pm y^3=(x\pm y)(x^2\mp xy+y^2)$
So when you are faced with $\sqrt[3]{x}$ you use that concept.

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