Preliminary Question on Fractions with a Rational Denominator

[Alyssapensini]

Hi all,

This is a prelim question. I should probably know this, but I just don't get it.
 
How do I rationalise the denominator and express as a single fraction?
I have attached the questions. Just label each one you explain as labelled in picture.

Thank you!
Alyssa

[jakesilove]

Hi all,

This is a prelim question. I should probably know this, but I just don't get it.
 
How do I rationalise the denominator and express as a single fraction?
I have attached the questions. Just label each one you explain as labelled in picture.

Thank you!
Alyssa

Hey Alyssa!

I've answered the first two of the questions you posted, because I think once you have a general idea of how it works you should be able to figure it out. Always take each fraction separately, and aim to rationalise the denominator. As will be explained, you basically do this by multiplying the top and the bottom of the fraction by the reverse of the denominator. Hope this helps!



Jake

[Happy Physics Land]

Hi all,

This is a prelim question. I should probably know this, but I just don't get it.
 
How do I rationalise the denominator and express as a single fraction?
I have attached the questions. Just label each one you explain as labelled in picture.

Thank you!
Alyssa

Hey Alyssa:

I find question (f) quite interesting and it is going to be a little difficult for you to solve at this stage since you havent yet done binomial theorem, which means that you would be required to expand a lot of brackets and this would be a tedious process. So I will post my worked solution here as well just in case you counter any difficulties.



But the main reason I decided to do question (f) is because that is a concept that's often used in extension 2 (usually in complex number proofs). z^2 - 1/z^2 would equal to 2i(sin2x) if we let z = cis x.

Sorry you wont need to know about anything that I am going to write below but I just cant resist my urge to prove this property:

Let z = cosx + isinx
z^n = (cosx + isinx)^n
z^n = cos(nx) + isin(nx)
1/z^n = z^-n
          = (cosx + isinx)^-n
          = cos(-nx) + isin(-nx)
          = cos (nx) - isin(nx) (Because cos x is an even function whereas sin x is an odd function)
Hence z^n - 1/z^n = cos (nx) + isin(nx) - (cos (nx) - isin(nx))
                             = 2isin(nx)

So in the case of z^2 - 1/z^2, it will equal to 2isin2x = 4isinxcosx, but anyways these are all non-relevant to you, I'm just posting this here to rejuvenate my own ego hehe.

Have fun studying!

Best Regards
Happy Physics Land

[jamonwindeyer]

Hey Alyssa! Awesome question with some awesome solutions.

However, I want to (probably to your great relief) say that the solution given by HPL is unnecessarily complicated. It uses the binomial theorem and some clever math, and it is correct, however: You definitely don't need any Extension knowledge to do it . And don't let the complex number concepts scare you, they don't apply to this question. However HPL, awesome explanation and some nice knowledge of Extension 2 course shown 

A much more familiar solution to question F is below:

================================







So we arrive at this line, and from here, we bring everything into one fraction:






And now we rationalise:





So the solution is:



I hope this helps Alyssa! If any steps here don't quite make sense, let me know and I'll lend a hand!

[Happy Physics Land]

Hey Alyssa! Awesome question with some awesome solutions.

However, I want to (probably to your great relief) say that the solution given by HPL is unnecessarily complicated. It uses the binomial theorem and some clever math, and it is correct, however: You definitely don't need any Extension knowledge to do it . And don't let the complex number concepts scare you, they don't apply to this question. However HPL, awesome explanation and some nice knowledge of Extension 2 course shown 

A much more familiar solution to question F is below:

================================







So we arrive at this line, and from here, we bring everything into one fraction:






And now we rationalise:





So the solution is:



I hope this helps Alyssa! If any steps here don't quite make sense, let me know and I'll lend a hand!

My ignorance towards the easier solutions ... just a part of the fatal flaw of extension students I guess

[jamonwindeyer]

My ignorance towards the easier solutions ... just a part of the fatal flaw of extension students I guess

Agreed. Sometimes you get pushed the hard stuff so much, you forget the simpler methods.