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#### wishmaster

##### Active member

- Oct 11, 2013

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**Re: Root calculations**

How would you isolate the radical in this example:

\(\displaystyle n(\sqrt{n^2+4)}-n^2\)

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- Oct 11, 2013

- 211

How would you isolate the radical in this example:

\(\displaystyle n(\sqrt{n^2+4)}-n^2\)

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It is hard to interpret the given expression since the opening bracket is outside of the radical and the closing bracket is inside. Do you simply mean:How would you isolate the radical in this example:

\(\displaystyle n(\sqrt{n^2+4)}-n^2\)

\(\displaystyle n\sqrt{n^2+4}-n^2\) ?

There is no need to enclose a radical with parentheses, as the radical itself serves as a grouping symbol on its own.

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- #3

- Oct 11, 2013

- 211

Im sure about the brackets.....it was so written in asignement.It is hard to interpret the given expression since the opening bracket is outside of the radical and the closing bracket is inside. Do you simply mean:

\(\displaystyle n\sqrt{n^2+4}-n^2\) ?

There is no need to enclose a radical with parentheses, as the radical itself serves as a grouping symbol on its own.

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- #4

It is a misprint then...let's write it instead in the more standard form:Im sure about the brackets.....it was so written in asignement.

\(\displaystyle n\sqrt{n^2+4}-n^2\)

I am thinking this comes from:

\(\displaystyle \lim_{n\to\infty}\left(n\sqrt{n^2+4}-n^2 \right)\)

Am I correct?

Given that we were discussing the rationalization of expressions, what you want to do is factor first...what factor is common to both terms in the expression?

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- #5

- Oct 11, 2013

- 211

Its n ?It is a misprint then...let's write it instead in the more standard form:

\(\displaystyle n\sqrt{n^2+4}-n^2\)

I am thinking this comes from:

\(\displaystyle \lim_{n\to\infty}\left(n\sqrt{n^2+4}-n^2 \right)\)

Am I correct?

Given that we were discussing the rationalization of expressions, what you want to do is factor first...what factor is common to both terms in the expression?

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- #6

Correct! So factoring this common factor out, what do you get?Its n ?

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- #7

- Oct 11, 2013

- 211

\(\displaystyle n(\sqrt{n^2+4}-n)\)Correct! So factoring this common factor out, what do you get?

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- #8

Correct again! You're on a roll!\(\displaystyle n(\sqrt{n^2+4}-n)\)

Now, if you are going to rationalize the numerator here*, with what do you think you should multiply this expression?

*Recall that \(\displaystyle f(n)=\frac{f(n)}{1}\).

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- #9

- Oct 11, 2013

- 211

With \(\displaystyle n(\sqrt{n^2+4}+n)\) ??Correct again! You're on a roll!

Now, if you are going to rationalize the numerator here*, with what do you think you should multiply this expression?

*Recall that \(\displaystyle f(n)=\frac{f(n)}{1}\).

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- #10

You could do that, but it is simpler to multiply by:With \(\displaystyle n(\sqrt{n^2+4}+n)\) ??

\(\displaystyle 1=\frac{\sqrt{n^2+4}+n}{\sqrt{n^2+4}+n}\)

I recommend that you try both and see that you get the same result, but with what you suggest, you have an extra step of reducing the result.

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- #11

- Oct 11, 2013

- 211

Can i see how you would do it? If you have time for me of course.......You could do that, but it is simpler to multiply by:

\(\displaystyle 1=\frac{\sqrt{n^2+4}+n}{\sqrt{n^2+4}+n}\)

I recommend that you try both and see that you get the same result, but with what you suggest, you have an extra step of reducing the result.

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- #13

- Oct 11, 2013

- 211

\(\displaystyle \frac{n^2+4-n^2}{\sqrt{n^2+4}+n}\)I would write:

\(\displaystyle n\left(\frac{\sqrt{n^2+4}-n}{1}\cdot\frac{\sqrt{n^2+4}+n}{\sqrt{n^2+4}+n} \right)\)

What do you get when you carry out the multiplication within the parentheses?

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- #14

Good! Now combine terms in the numerator, and don't forget the factor of $n$ that was outside of the parentheses. What do you wind up with?\(\displaystyle \frac{n^2+4-n^2}{\sqrt{n^2+4}+n}\)

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- #15

- Oct 11, 2013

- 211

\(\displaystyle \frac{4n}{\sqrt{n^2+4}+n}\)Good! Now combine terms in the numerator, and don't forget the factor of $n$ that was outside of the parentheses. What do you wind up with?

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- #16

Excellent! Now, what were you wanting to do with this expression?\(\displaystyle \frac{4n}{\sqrt{n^2+4}+n}\)

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- #17

- Oct 11, 2013

- 211

I wanted just to simplify it.......Excellent! Now, what were you wanting to do with this expression?

But i believe,thast not ok for my problem with limit as you mentioned...or?

I think i am improving in math with your help,but i still dont get the logic of math....to do right steps and so....

What are you looking at first? To isolate the radical? To prove what? This are my problems now!

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- #18

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- #19

- Oct 11, 2013

- 211

If, as I suspect, you are wanting to take the limit of this expression as $n\to\infty$, then you want to divide each term in the numerator and denominator by $n$.

Can i continue from that last expression?

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- #20

Yes, that is the expression that I meant when dividing each term by $n$.Can i continue from that last expression?

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- #21

- Oct 11, 2013

- 211

I still dot get it whats the most important thing to do with radicals......how to "fight" with them.....Yes, that is the expression that I meant when dividing each term by $n$.

I mean,as a question of limits,whats the most important thing to do....?

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- #22

As far as limits go, you want to get a determinate form. Forms such as $\infty-\infty$ or $\dfrac{\infty}{\infty}$ are indeterminate. But when you rationalize the numerator in the case of your problem and then divide each term by $n$ you get a determinate form.I still dot get it whats the most important thing to do with radicals......how to "fight" with them.....

I mean,as a question of limits,whats the most important thing to do....?

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- #23

- Oct 11, 2013

- 211

Silly question...what is determinate form?As far as limits go, you want to get a determinate form. Forms such as $\infty-\infty$ or $\dfrac{\infty}{\infty}$ are indeterminate. But when you rationalize the numerator in the case of your problem and then divide each term by $n$ you get a determinate form.

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- #24

A determinate form is one which you can find the limit by substitution of the limiting value for the variable. If you try this with the original form you gave, you get $\infty-\infty$ and you cannot determine anything from this. However, in the final form (after you divide each term by $n$) you find a real number, because it is now a determinate form.Silly question...what is determinate form?

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- #25

- Oct 11, 2013

- 211

So i believe this is a thing i should learn.........question is how....A determinate form is one which you can find the limit by substitution of the limiting value for the variable. If you try this with the original form you gave, you get $\infty-\infty$ and you cannot determine anything from this. However, in the final form (after you divide each term by $n$) you find a real number, because it is now a determinate form.