MHB Root Calculation: Isolating the Radical

[theakdad]

Re: Root calculations

How would you isolate the radical in this example:

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

 

[MarkFL]

Re: Root calculations

How would you isolate the radical in this example:

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

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:

$$n\sqrt{n^2+4}-n^2$$ ?

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

 

[theakdad]

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:

$$n\sqrt{n^2+4}-n^2$$ ?

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]

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:

$$n\sqrt{n^2+4}-n^2$$

I am thinking this comes from:

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

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?

 

[theakdad]

Re: Root calculations

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

$$n\sqrt{n^2+4}-n^2$$

I am thinking this comes from:

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

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]

Re: Root calculations

Its n ?

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

 

[theakdad]

Re: Root calculations

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

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

 

[MarkFL]

Re: Root calculations

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

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 $$f(n)=\frac{f(n)}{1}$$.

 

[theakdad]

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 $$f(n)=\frac{f(n)}{1}$$.

With $$n(\sqrt{n^2+4}+n)$$ ??

 

[MarkFL]

Re: Root calculations

With $$n(\sqrt{n^2+4}+n)$$ ??

You could do that, but it is simpler to multiply by:

$$1=\frac{\sqrt{n^2+4}+n}{\sqrt{n^2+4}+n}$$

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

 

[theakdad]

Re: Root calculations

You could do that, but it is simpler to multiply by:

$$1=\frac{\sqrt{n^2+4}+n}{\sqrt{n^2+4}+n}$$

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]

I would write:

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

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

 

[theakdad]

I would write:

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

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

$$\frac{n^2+4-n^2}{\sqrt{n^2+4}+n}$$

 

[MarkFL]

$$\frac{n^2+4-n^2}{\sqrt{n^2+4}+n}$$

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?

 

[theakdad]

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?

$$\frac{4n}{\sqrt{n^2+4}+n}$$

 

[MarkFL]

$$\frac{4n}{\sqrt{n^2+4}+n}$$

Excellent! Now, what were you wanting to do with this expression?

 

[theakdad]

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 don't 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]

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 $n$.

 

[theakdad]

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]

Can i continue from that last expression?

Yes, that is the expression that I meant when dividing each term by $n$.

 

[theakdad]

Yes, that is the expression that I meant when dividing each term by $n$.

I still dot get it what's 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]

I still dot get it what's 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.

 

[theakdad]

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]

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.

 

[theakdad]

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...

 

[MarkFL]

So i believe this is a thing i should learn...question is how...

You are on the path...I have noticed a good improvement in your algebra since you have been here. (Yes)

Just keep at it, and gradually the pieces come together. :D

 

[theakdad]

You are on the path...I have noticed a good improvement in your algebra since you have been here. (Yes)

Just keep at it, and gradually the pieces come together. :D

Yes,i have improved...with your help,and i apreciate this,and I am very thankfull to MHB!

Can you just tell me more about determinate form? What exactly does it menas?

 

[MarkFL]

Consider the following limit:

$$\lim_{x\to 9}\frac{x-9}{\sqrt{x}-3}$$

If we try to substitute $9$ for $x$, we get $$\frac{0}{0}$$, which is an indeterminate form, as it tells us nothing about the actual value of the limit...it could be anything as far as we know at this point. However, if we factor:

$$\lim_{x\to 9}\frac{\left(\sqrt{x}+3 \right)\left(\sqrt{x}-3 \right)}{\sqrt{x}-3}$$

And now reduce:

$$\lim_{x\to 9}\left(\sqrt{x}+3 \right)$$

Now when we make the substitution, we get a real number:

$$\lim_{x\to 9}\left(\sqrt{x}+3 \right)=\sqrt{9}+3=6$$

This is because we transformed the expression into a determinate form, that is, we can determine the value of the limit by a simple substitution.

 

[theakdad]

Consider the following limit:

$$\lim_{x\to 9}\frac{x-9}{\sqrt{x}-3}$$

If we try to substitute $9$ for $x$, we get $$\frac{0}{0}$$, which is an indeterminate form, as it tells us nothing about the actual value of the limit...it could be anything as far as we know at this point. However, if we factor:

$$\lim_{x\to 9}\frac{\left(\sqrt{x}+3 \right)\left(\sqrt{x}-3 \right)}{\sqrt{x}-3}$$

And now reduce:

$$\lim_{x\to 9}\left(\sqrt{x}+3 \right)$$

Now when we make the substitution, we get a real number:

$$\lim_{x\to 9}\left(\sqrt{x}+3 \right)=\sqrt{9}+3=6$$

This is because we transformed the expression into a determinate form, that is, we can determine the value of the limit by a simple substitution.

Thank you for the explanation...
Soon we move into calculus,im wondering if i would be able to deal with it...