Maths year 11

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    (Original post by RDKGames)
    Numerator's incorrect.
    How?

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    Sign is wrong.
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    (Original post by RDKGames)
    Sign is wrong.


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    Correct.
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    Can you rationalise  \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.
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    (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 3\sqrt2 and 2\sqrt2 because they share a common factor.
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    (Original post by RDKGames)
    Correct. Though remember you can add 3\sqrt2 and 2\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 3\sqrt2 + 2\sqrt2 and you can factor out \sqrt2 out of them both to get \sqrt2(3+2) and you can add the 3 and 2 together inside the bracket to get a 5; hence giving \sqrt2(5)=5\sqrt2
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    (Original post by RDKGames)
    You have 3\sqrt2 + 2\sqrt2 and you can factor out \sqrt2 out of them both to get \sqrt2(3+2) and you can add the 3 and 2 together inside the bracket to get a 5; hence giving \sqrt2(5)=5\sqrt2


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    So you have  3\sqrt 2-\sqrt 4+6-2\sqrt 2 . Now  \sqrt 4 =2 and  3\sqrt 2 -2\sqrt 2=\sqrt 2 . Use all this and simplify, you went a bit off after this line in your workings.
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    (Original post by RDKGames)
    You have 3\sqrt2 + 2\sqrt2 and you can factor out \sqrt2 out of them both to get \sqrt2(3+2) and you can add the 3 and 2 together inside the bracket to get a 5; hence giving \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 3\sqrt2 + 2\sqrt2 comes from. And I'm not sure how you got -6\sqrt4 from (-\sqrt4 + 6)
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    (Original post by B_9710)
    So you have  3\sqrt 2-\sqrt 4+6-2\sqrt 2 . Now  \sqrt 4 =2 and  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  \displaystyle \frac{47}{2+\sqrt 3+\sqrt 6} ?
    No I cannot rationalise that.
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    But I can rationalise the denominator:
    \displaystyle \sqrt6 + 7\sqrt3 - 12\sqrt2 + 10
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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 \sqrt2 when you supposed to subtract one from the other. Secondly 6-\sqrt4 is not -6\sqrt4. It's 4.
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    (Original post by RDKGames)
    Indeed you are getting confused. Try to follow what you are doing. Firstly you added the terms with \sqrt2 when you supposed to subtract one from the other. Secondly 6-\sqrt4 is not -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.
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    (Original post by RDKGames)
    Yep, except the last term.
    Now



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    Good.
 
 
 
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