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Thread: Binomial Expansion-can you help solve this question?

  1. #1
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    Binomial Expansion-can you help solve this question?

    Can anyone help solve the first question? I get an answer of 89 - 28(2^1/2)(5^1/2) but I don't see how to get in the form required where a and b are integers. I can only get a and b as surds. Help!

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    Re: Binomial Expansion-can you help solve this question?

    Quote Originally Posted by zndavies View Post
    Can anyone help solve the first question? I get an answer of 89 - 28(2^1/2)(5^1/2) but I don't see how to get in the form required where a and b are integers. I can only get a and b as surds. Help!
    You must do some work of your own.

    For the first: $(a+b)^4=\dbinom{4}{0}a^4b^{4-4}+\dbinom{4}{1}a^3b^{4-3}+\dbinom{4}{2}a^2b^{4-2}+\dbinom{4}{3}a^1b^{4-1}+\dbinom{4}{4}a^0b^{4-0}$
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    Re: Binomial Expansion-can you help solve this question?

    Have you managed to get the answer in the specified form? I know what the binomial expansion is and I get an answer but not in the form they are asking for.
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    Re: Binomial Expansion-can you help solve this question?

    Quote Originally Posted by zndavies View Post
    Have you managed to get the answer in the specified form? I know what the binomial expansion is and I get an answer but not in the form they are asking for.
    Then take one of the terms:
    $ \begin{align*} \dbinom{4}{3}a^1b^{4-1}&=\dfrac{4!}{1!\cdot3!}(\sqrt{5}^1)(-\sqrt{6})^3\\&=\dfrac{4!}{1!\cdot3!}(\sqrt{5})(-6\sqrt{6})\\&=\dfrac{4!}{1!\cdot3!}(-6\sqrt{30})\end{align*}$

    Now can you do elementary arithmetic?
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    Re: Binomial Expansion-can you help solve this question?

    $\displaystyle \left( \sqrt{5}-\sqrt{2} \right) ^4=a\sqrt{5}+b\sqrt{2}$

    does not seem to have solutions in integers a and b

    on the other hand

    $\displaystyle \left( \sqrt{5}-\sqrt{2} \right) ^3=-17 \sqrt{2}+11 \sqrt{5}$
    Last edited by Idea; Apr 14th 2018 at 08:04 AM.
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    Re: Binomial Expansion-can you help solve this question?

    Quote Originally Posted by Idea View Post
    $\displaystyle \left( \sqrt{5}-\sqrt{2} \right) ^4=a\sqrt{5}+b\sqrt{2}$
    does not seem to have solutions in integers a and b
    Note that the question did not ask for solutions of $a~\&~b$ but rather asks for answers in the form of $a\sqrt{5}+b\sqrt{2}$
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    Re: Binomial Expansion-can you help solve this question?

    Yes but it is asking for it in specifically in that form with a and b integers 'to be found'?
    I have a degree in maths and like Idea I do not think there is such an answer where a and b are integers.
    I posted on here in case I was wrong to see if anyone could find an answer, didn't expect for people to basically be rude and ask me if I can do elementary arithmetic.
    You still haven't answered whether or not you can find an answer in the correct form which I assume means you can't.
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    Re: Binomial Expansion-can you help solve this question?

    Quote Originally Posted by zndavies View Post
    ..didn't expect for people to basically be rude and ask me if I can do elementary arithmetic.
    Take that with a grain of salt: Plato is a smart clown who enjoys "faking" insults;
    really doesn't mean 'em: stick around, you'll get to know Mr. Grouch
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    Re: Binomial Expansion-can you help solve this question?

    Quote Originally Posted by zndavies View Post
    Yes but it is asking for it in specifically in that form with a and b integers 'to be found'?
    I have a degree in maths and like Idea I do not think there is such an answer where a and b are integers.
    I posted on here in case I was wrong to see if anyone could find an answer, didn't expect for people to basically be rude and ask me if I can do elementary arithmetic.
    You still haven't answered whether or not you can find an answer in the correct form which I assume means you can't.
    4) $(\sqrt{5}-\sqrt{2})^4=89-28\sqrt{10}$
    See here.

    5a.) $(1+\sqrt{6})^5=141+101\sqrt{6}$
    See HERE
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    Re: Binomial Expansion-can you help solve this question?

    5b) n=11
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    Re: Binomial Expansion-can you help solve this question?

    Thanks guys, I was only looking to an answer to Q4 and Plato you got the same answer as me.
    From this I'm going to assume this question is 'wrong' as I thought in that it isn't possible to find the answer in the required form (not with a and b integers anyway).
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    Re: Binomial Expansion-can you help solve this question?

    Quote Originally Posted by zndavies View Post
    You still haven't answered whether or not you can find an answer in the correct form which I assume means you can't.
    Here is 5b. worked out.
    The term in the expansion of $(1+3x)^n$ which contains $x^2$ looks like:
    $ \begin{align*}\dbinom{n}{2}(1)^{n-2}(3x)^2&=\dfrac{n!}{2!(n-2)!}(9x^2) \\&=\dfrac{ n(n-1)}{2!}(9x^2) \end{align*}$

    So now $ \begin{align*}\dfrac{ n(n-1)}{2!}(9) &=\dfrac{9n(n-1)}{2}\\&=495 \\9n(n-1)&=2\cdot 995\\n^2-n-110&=0\\(n-11)(n+10)&=0\\n&=11 \end{align*}$

    Check it out HERE.
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