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Maths year 11

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Reply 1240
Original post by RDKGames
Numerator's incorrect.


How?

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Sign is wrong.
Reply 1242
Original post by RDKGames
Sign is wrong.




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Can you rationalise 472+3+6 \displaystyle \frac{47}{2+\sqrt 3+\sqrt 6} ?
Original post by RDKGames
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.
(edited 7 years ago)
Reply 1245
Original post by RDKGames
Correct.


Wb this?

Thank you so much! !!!

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Original post by z_o_e
Wb this?

Thank you so much! !!!

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Correct. Though remember you can add 323\sqrt2 and 222\sqrt2 because they share a common factor.
Reply 1247
Original post by RDKGames
Correct. Though remember you can add 323\sqrt2 and 222\sqrt2 because they share a common factor.


I'm not sure about this one.


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Original post by z_o_e
I'm not sure about this one.


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You have 32+223\sqrt2 + 2\sqrt2 and you can factor out 2\sqrt2 out of them both to get 2(3+2)\sqrt2(3+2) and you can add the 3 and 2 together inside the bracket to get a 5; hence giving 2(5)=52\sqrt2(5)=5\sqrt2
Reply 1249
Original post by RDKGames
You have 32+223\sqrt2 + 2\sqrt2 and you can factor out 2\sqrt2 out of them both to get 2(3+2)\sqrt2(3+2) and you can add the 3 and 2 together inside the bracket to get a 5; hence giving 2(5)=52\sqrt2(5)=5\sqrt2




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So you have 324+622 3\sqrt 2-\sqrt 4+6-2\sqrt 2 . Now 4=2 \sqrt 4 =2 and 3222=2 3\sqrt 2 -2\sqrt 2=\sqrt 2 . Use all this and simplify, you went a bit off after this line in your workings.
Reply 1251
Original post by RDKGames
You have 32+223\sqrt2 + 2\sqrt2 and you can factor out 2\sqrt2 out of them both to get 2(3+2)\sqrt2(3+2) and you can add the 3 and 2 together inside the bracket to get a 5; hence giving 2(5)=52\sqrt2(5)=5\sqrt2




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So much wrong in that. You expanded right but collecting the terms is the thing that trips you up. Like I don't know where the 32+223\sqrt2 + 2\sqrt2 comes from. And I'm not sure how you got 64-6\sqrt4 from (4+6)(-\sqrt4 + 6)
Reply 1253
Original post by B_9710
So you have 324+622 3\sqrt 2-\sqrt 4+6-2\sqrt 2 . Now 4=2 \sqrt 4 =2 and 3222=2 3\sqrt 2 -2\sqrt 2=\sqrt 2 . Use all this and simplify, you went a bit off after this line in your workings.


This whole things confused me... starting from these red dots...

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Original post by Ano123
Can you rationalise 472+3+6 \displaystyle \frac{47}{2+\sqrt 3+\sqrt 6} ?


No I cannot rationalise that.

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Original post by z_o_e
This whole things confused me... starting from these red dots...

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Indeed you are getting confused. Try to follow what you are doing. Firstly you added the terms with 2\sqrt2 when you supposed to subtract one from the other. Secondly 646-\sqrt4 is not 64-6\sqrt4. It's 4.
Reply 1256
Original post by RDKGames
Indeed you are getting confused. Try to follow what you are doing. Firstly you added the terms with 2\sqrt2 when you supposed to subtract one from the other. Secondly 646-\sqrt4 is not 64-6\sqrt4. It's 4.


Yeah is this part correct though?



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Original post by z_o_e
Yeah is this part correct though?



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Yep, except the last term.
Reply 1258
Original post by RDKGames
Yep, except the last term.

Now



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