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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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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$\displaystyle \left( \sqrt{5}-\sqrt{2} \right) ^4=a\sqrt{5}+b\sqrt{2}$
does not seem to have solutions in integers a and b
on the other hand
$\displaystyle \left( \sqrt{5}-\sqrt{2} \right) ^3=-17 \sqrt{2}+11 \sqrt{5}$
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.
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).
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.