Maths year 11

I don't understand this.

Can you do this as one question as an example cause I've got others to do like similar ones.

I started rationalising.



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You can simplify q4 and q5 further.

Example of the question (bottom part of the page):




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Thank you so for 5 and 6 you mean the surd 6 can be simplified ?

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No that surd cannot be simplified any further. But the fact that you have terms of 12/6 and 4/2, those can be.

Why are you dissecting root 6...? I said that you cannot simplify that any further.

Yeah I fixed this.


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Well your answer is right because you've made two wrong's which cancel each other out. For the first line you still made the same mistake but with opposite sign now.

I give up.


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The problem here is that it doesn't look like you know what you're doing and why you're doing it. Simply guessing hoping to get the right answer.

But I looked at your example and followed it.


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But do you understand the maths in it?

Yes

Could you point out where the mistake was so I could fix it in green pen and not make that on the other questions. I'd rather keep on going cause I made it this far.

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Denominator after the equals sign on line 1: should be $(\sqrt5 - 1)(\sqrt5 + 1)$. The rest is fine.

If you understand it, explain to me why we multiply $\frac{1}{\sqrt5 - 1}$ by $\frac{\sqrt5 + 1}{\sqrt5 + 1}$ rather than $\frac{\sqrt5 - 1}{\sqrt5 - 1}$ as you would initially guess.

Cause we do that rule (a+b) (a-b)

And then do the double brackets and do the squares and work that out to find the denominator of the answer.

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That's not the reason. What you've said applies to either case hence it's not the reason.

What is important about (a+b)(a-b) is that it is the difference of two squares whereby (a+b)(a-b)=a²-b². So when either a or b is a surd, it will always become an integer. The problem with (a+b)(a+b) is that it equals a²+b²+2ab and due to the 2ab term we would never rationalise the surds, hence never rationalise the denominator.

how did this go?


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