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For my exercise, I have been asked to rationalize and simplify this surd;

$$\frac{x-1}{\sqrt{2\sqrt{x}} + 1 - \sqrt[4]{x}}$$

I don't know how to type it. The denominator is square root of 2 with square root of x after that is + 1 - fourth root of x.]

Each time I do this I get the wrong answer. The method I am using is;

  1133√−7×33√−733√−7

I'm confused with the $\sqrt{2}$,where $\sqrt{x}$ is inside the $\sqrt{2}$ and $1$ separated from $\sqrt{2}$ and $\sqrt{x}$. Sorry if I can't really elaborate it correctly.

This ends up nowhere near the right answer, even once it is simplified, can someone tell me where i'm going wrong?

Many thanks!

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2  
You can use $\TeX$ on this site, enclosed in single dollar signs for inline formulas or in double dollar signs for displayed equations. The $\TeX$ command for square roots is \sqrt. The way you're writing the square roots makes them very hard to read. –  joriki Jul 17 '12 at 13:21
    
Let's get the fraction straightened out too. Is it $\frac{x-1}{\sqrt{2x}+1-\sqrt[4]{x}}$ or $\frac{x-1}{\sqrt{2x}+1}-\sqrt[4]{x}$? Or is it something else. –  peoplepower Jul 17 '12 at 13:26
    
x−12x√+1−x√4 sir something like that sir but the √x is inside of √2. –  user12515 Jul 17 '12 at 13:30
    
Ah,and is the $\sqrt[4]{x}$ outside or inside the denominator? –  peoplepower Jul 17 '12 at 13:31
1  
@user12515: I had explained how to type it. It would help if you could explain what part of the explanation you didn't understand, instead of just throwing your hands up and saying that you don't know how to type it. Perhaps the missing part in the explanation was that you need to put curly braces around radicands consisting of more than one character? E.g. \sqrt{2x} produces $\sqrt{2x}$. –  joriki Jul 17 '12 at 13:33

4 Answers 4

Note that $\sqrt{2\sqrt x}=\sqrt2\root4\of x$, so your denominator (if I have understood your comments) is $1+(\sqrt2-1)\root4\of x$. Can you rationalize now?

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This is confusing as in the question it is clearly written $\,\sqrt 2\,\sqrt x\,$ , and not $\,\sqrt {2\sqrt x}\,$ , and until the OP doesn't learn the very little LaTeX needed to write decently mathematics in ME we, apparently, won't even be sure what he meant to write. –  DonAntonio Jul 17 '12 at 13:37
    
@Don: The $\sqrt2 \sqrt x$ was edited in by someone else, presumably as an interpretation of "The denominator is square root of 2 with square root of x". –  joriki Jul 17 '12 at 13:46
    
Sorry sir i'm just trying to say it in English language sir. I'm not really meant to speak in English because this is not my language sir. I have language barrier. SORRY again. I'm just a student trying to LEARN. But Thank you because you corrected me. THANK YOU :) –  user12515 Jul 17 '12 at 13:46
    
@user12515: First, it is not a language barrier but a writing-mathematics barrier. Second, you can try to write in your language and, if you're lucky, perhaps somebody understands it and can help you. Anyway, in your own language or in english, you must learn how to write mathematics in this site, otherwise it can be hard, or simply impossilbe, to understand what you meant to write. –  DonAntonio Jul 17 '12 at 13:50
    
@DonAntonio Ok sir I will start to study how to write in mathematics in this site Sir for me to express what I meant to write. A BIG THANKS to You Sir. I learned a lot with your comments sir. –  user12515 Jul 17 '12 at 14:02

It seems to me that there are (at least) two possible problems we can consider. The first problem is to rewrite the expression with a polynomial denominator, and the second problem is to rewrite the expression with an integer-coefficient polynomial denominator. I'll first take care of the first problem and then continue so as to take care of the second problem.

$$\frac{x-1}{\sqrt{2\sqrt x\;}+1-\sqrt[4] x}\;\;=\;\;\frac{x-1}{1+(\sqrt 2-1)\sqrt[4] x}\;\;=\;\;\frac{x-1}{1-(1-\sqrt 2)\sqrt[4] x}$$

Using the identity $1-u^4=(1-u)\left(1+u+u^2+u^3\right)$ with $u=(1-\sqrt 2)\sqrt[4] x,$ we can multiply both the numerator and the denominator of the right-most displayed fraction above to get

$$\frac{(x-1)\left(1+u+u^2+u^3\right)}{1-(1-\sqrt 2)^4x},$$

which takes care of the first problem. To solve the second problem, first note that the last fraction is equal to

$$\frac{(x-1)\left(1+u+u^2+u^3\right)}{(1+17x)-12x\sqrt{2}}$$

Now simply multiply both the numerator and the denominator by $(1+17x)+12x\sqrt{2},$ and we're done.

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As Gerry wrote in his answer:

$$\frac{x-1}{\sqrt{2\sqrt x}+1-\sqrt[4] x}=\frac{x-1}{1+(\sqrt 2-1)\sqrt[4] x}=\frac{x}{1+(\sqrt 2-1)\sqrt[4] x}-\frac{1}{1+(\sqrt 2-1)\sqrt[4] x}$$ Can you take it from here?

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@DonAtonio is this the time I can rationalize the denominator Sir? –  user12515 Jul 17 '12 at 14:28
1  
Er...well, you can try to rationalize fractions, or expressions in general, but this could be a language barrier: what exactly do you understand by "rationalize" something? –  DonAntonio Jul 17 '12 at 14:50

Hint $ $ Let $\rm\: z\, =\, (1\!-\!\sqrt 2)\sqrt[4]x,\ $ so $\rm\:z^2 = (3\!-\!2\sqrt2)\sqrt x,\,$ $\rm\:z^4\, =\, (17\!-\!12\sqrt2)\,x\, =:\, \alpha\,x,\ $ hence

$$\begin{eqnarray}\rm \frac{1}{1-(1\!-\!\sqrt 2)\sqrt[4]x} &\ =\ &\rm \frac{1}{1-z}\ =\ \frac{1+z\,\ }{1-z^2}\ =\ \frac{(1+z)\,(1+z^2)}{1- z^4}\\ &\ =\ &\rm \frac{(1+z)\,(1+z^2)}{1-\alpha \,x}\ =\ \frac{(1+z)\,(1+z^2)(1-\alpha'x)}{1-(\alpha\!+\!\alpha')\,x+\alpha\alpha'\,x^2}\end{eqnarray} $$

In our case we have $\rm\, \alpha\!+\!\alpha' = 34,\ \alpha\alpha' = 1,\,$ so the denominator is $\rm\:1 - 34\,x + x^2.$ The final result is obtained by multiplying by $\rm\:x\!-\!1\:$ and substituting in the above the actual values of $\rm\:z\:$ and $\rm\:z^2.$

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