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

1. ## 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!

Sent from my SM-G925F using Tapatalk

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

Originally Posted by zndavies
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: $(a+b)^4=\dbinom{4}{0}a^4b^{4-4}+\dbinom{4}{1}a^3b^{4-3}+\dbinom{4}{2}a^2b^{4-2}+\dbinom{4}{3}a^1b^{4-1}+\dbinom{4}{4}a^0b^{4-0}$

3. ## 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.

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

Originally Posted by zndavies
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?

5. ## 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

$\displaystyle \left( \sqrt{5}-\sqrt{2} \right) ^3=-17 \sqrt{2}+11 \sqrt{5}$

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

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

7. ## 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.

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

Originally Posted by zndavies
..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

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

Originally Posted by zndavies
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

5b) n=11

11. ## 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).

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

Originally Posted by zndavies
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.