• June 21st 2012, 02:00 AM
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?

• June 21st 2012, 02:19 AM
Goku
$\frac{1}{h}(\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt{x}})$

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

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

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

Now continue from there....(Evilgrin)
• June 21st 2012, 02:41 AM
biffboy
Quote:

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?

Multiply numerator and denominator by root(x+h)root(x) Then multiply numerator and denominator by root(x)+root(x+h)
• June 21st 2012, 02:56 AM
SVDM
Hallo ou maat!

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

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

Stan
• June 21st 2012, 03:51 AM
Goku
Remember that $a^2 - b^2 = (a-b)(a+b)$...
• June 21st 2012, 12:03 PM
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...

Attachment 24131

? I tried to solve this but the answer is so wrong...
• June 21st 2012, 12:16 PM
Plato
Quote:

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...

Attachment 24131

Multiply by numerator and denominator by $\left( {\sqrt[3]{{{x^2}}} - \sqrt[3]{x}\sqrt[3]{a} + \sqrt[3]{{{a^2}}}} \right)$
• June 21st 2012, 12:19 PM
SVDM
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.
• June 21st 2012, 02:04 PM
Reckoner
Quote:

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...

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

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

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

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

$=\frac1{\sqrt{x+1} + \sqrt{a+1}}$
• June 22nd 2012, 03:31 PM
SVDM
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
• June 22nd 2012, 06:53 PM
skeeter
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?
• June 22nd 2012, 08:39 PM
SVDM
Hi thanks,

So I did misunderstand Pluto above;
Quote:

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.
• June 23rd 2012, 02:26 PM
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.

Thanks again.
• June 23rd 2012, 02:39 PM
Plato
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.