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Isolating a radical

wishmaster

Active member
Oct 11, 2013
211
Re: Root calculations

How would you isolate the radical in this example:

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

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: Root calculations

How would you isolate the radical in this example:

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

wishmaster

Active member
Oct 11, 2013
211
Re: Root calculations

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.
Im sure about the brackets.....it was so written in asignement.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: Root calculations

Im sure about the brackets.....it was so written in asignement.
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?
 

wishmaster

Active member
Oct 11, 2013
211
Re: Root calculations

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?
Its n ?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775

wishmaster

Active member
Oct 11, 2013
211
Re: Root calculations

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

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: Root calculations

\(\displaystyle n(\sqrt{n^2+4}-n)\)
Correct again! (Star) 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 \(\displaystyle f(n)=\frac{f(n)}{1}\).
 

wishmaster

Active member
Oct 11, 2013
211
Re: Root calculations

Correct again! (Star) 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 \(\displaystyle f(n)=\frac{f(n)}{1}\).
With \(\displaystyle n(\sqrt{n^2+4}+n)\) ??
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: Root calculations

With \(\displaystyle n(\sqrt{n^2+4}+n)\) ??
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. :D
 

wishmaster

Active member
Oct 11, 2013
211
Re: Root calculations

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. :D
Can i see how you would do it? If you have time for me of course.......
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
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?
 

wishmaster

Active member
Oct 11, 2013
211
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?
\(\displaystyle \frac{n^2+4-n^2}{\sqrt{n^2+4}+n}\)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
\(\displaystyle \frac{n^2+4-n^2}{\sqrt{n^2+4}+n}\)
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?
 

wishmaster

Active member
Oct 11, 2013
211
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?
\(\displaystyle \frac{4n}{\sqrt{n^2+4}+n}\)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775

wishmaster

Active member
Oct 11, 2013
211
Excellent! Now, what were you wanting to do with this expression?
I wanted just to simplify it.......
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! :p
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
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 $n$.
 

wishmaster

Active member
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?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775

wishmaster

Active member
Oct 11, 2013
211
Yes, that is the expression that I meant when dividing each term by $n$.
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....?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
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....?
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.
 

wishmaster

Active member
Oct 11, 2013
211
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.
Silly question...what is determinate form?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Silly question...what is determinate form?
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.
 

wishmaster

Active member
Oct 11, 2013
211
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.
So i believe this is a thing i should learn.........question is how....