anonymous
  • anonymous
Expand and simplify (sqrt 3 + 2)^5. Give your answer in the form a+bsqrt
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
\[(\sqrt{3} + 2)^{5}\]
anonymous
  • anonymous
I know pascals triangle is going to make this 1, 5, 10, 5, 1 and how to order all the exponents and such. my work ended up looking like the following
anonymous
  • anonymous
\[\sqrt{3}^{5} + 5(\sqrt{3})^{4}(2)^{2} + 10(\sqrt{3})^3(2)^2 + 10(\sqrt{3})^2(2)^3 + 5(\sqrt{3})(2)^4 + (2)^5\]

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anonymous
  • anonymous
@freckles @whpalmer4 @UsukiDoll any of you guys awake and in the mood to (possibly) help out? I understand if not, it's p late ;-;
anonymous
  • anonymous
hi so um the answer in the back of the book is \[362+209\sqrt{3}\] but i got
anonymous
  • anonymous
\[521+1562\] \[521+1562\sqrt{3}\]
anonymous
  • anonymous
(the last one, the site just glitched and kept the first answer)
anonymous
  • anonymous
ahh, is it too difficult/long ;-;?
whpalmer4
  • whpalmer4
\[(a+b)^5 = a^5b^0 + a^4b^1+a^3b^2 + a^2b^3 + a^1b^4+a^0b^5\]with the appropriate coefficients, right? Why don't we simplify this part first? \[a^5 = \sqrt{3}^5 = 9\sqrt{5}\]\[a^4 = 9\]\[a^3=3\sqrt{3}\]\[a^2 = 3\]\[a^1 = \sqrt{3}\] \[b^5 = 32\]\[b^4=16\]\[b^3=8\]\[b^2=4\]\[b^1=2\] so keeping our term places together so we can stick in the coefficients, that's \[9\sqrt{3}*1 + 9*2+3\sqrt{3}*4 + 3*8 + \sqrt{3}*16+32\]and sticking in the coefficients from the triangle \[=1*9\sqrt{3}*1 + 5*9*2 + 10*3\sqrt{3}*4+10*3*8 + 5*\sqrt{3}*16+1*32\] \[=9\sqrt{3} + 90 + 120\sqrt{3} +240+80\sqrt{3}+32 = (9+120+80)\sqrt{3} + 90+240+32\]\[=362+209\sqrt{3} \]
anonymous
  • anonymous
ah man thank you so much!! makes sense now that I look at it. do you think you could help me with umm one more?
whpalmer4
  • whpalmer4
I am a huge fan of doing this kind of tedious work with simple things like \(a\) instead of \(\sqrt{3}\)! Sure, I could probably stumble through another one...
anonymous
  • anonymous
ok haha thank you. so basically, find the coefficient of \[x^-6\] in the expansion of \[(2x-\frac{ 3 }{ x^2})\]
anonymous
  • anonymous
this one just messes me up cos of the negative exponent
whpalmer4
  • whpalmer4
Isn't there supposed to be an exponent on that?
anonymous
  • anonymous
oh yea haha it's 12
anonymous
  • anonymous
could you explain this the way you did with the last example?
whpalmer4
  • whpalmer4
darn. that was looking pretty easy :-)
anonymous
  • anonymous
hehe sorry
whpalmer4
  • whpalmer4
Okay, well, we have to figure out the coefficients from the triangle. Can you do that? \[a^{12}b^0 ,a^{11}b^1, a^{10}b^2, a^9b^3, a^8b^4, a^7b^5 ,a^6b^6, a^5b^7, a^4b^8 ,a^3b^9, a^2b^{10} ,a^1b^{11} ,a^0b^{12}\]
whpalmer4
  • whpalmer4
Now we have \[a= 2x\]\[b=-3x^{-2}\]Right? And we want to find the coefficient of \(x^{-6}\) (by the way, if your exponent takes up multiple characters, like "-6", wrap the entire exponent in { } so that the right thing happens)
anonymous
  • anonymous
do you want me to do pascals triangle for 12?
whpalmer4
  • whpalmer4
Yes, please. Division of labor :-)
anonymous
  • anonymous
lol kk one sec
anonymous
  • anonymous
1 12 66 220 495 792 924 792 495 220 66 12 1
whpalmer4
  • whpalmer4
there's actually a formula that you can use to find the value of any of the coefficients, but I don't ever remember it!
anonymous
  • anonymous
i don't know it either :/
whpalmer4
  • whpalmer4
Yes, those the ones. Now can you tell me which term is going to have the \(x^{-6}\) in it?
anonymous
  • anonymous
umm...792?
anonymous
  • anonymous
is it the second 792?
whpalmer4
  • whpalmer4
well, in terms of a^n b^m by the way, the formula for it is\[\left(\begin{matrix}n \\ k\end{matrix}\right)= \frac{n!}{k!(n-k)!}\]
anonymous
  • anonymous
oh yaaa the binomial theorem
anonymous
  • anonymous
but how can that be used to find the coefficient here? can we do so by finding the general term?
whpalmer4
  • whpalmer4
The first term is \(a^{12}b^0\) which is going to \(x^{12}*1\) The second term is \(a^{11}b^1\) which is going to have \(x^{11}*x^{-2} = x^9\) The third term is \(a^{10}b^2\) which is going to have \(x^{10}*x^{-2*2} = x^6\) so the exponent is decreasing by 3 with each term...that means x^3, x^0, x^-3,x^-6 7th term is the one we want that will be \(a^6b^6 = (2x)^6(-3x^-2)^6 = \) times our coefficient which is 924
anonymous
  • anonymous
ok
whpalmer4
  • whpalmer4
oops, I muffed some formatting. 7th term is \(924a^6b^6 = 924(2x)^6(-3x^{-2})^6 = \) can you do that part?
anonymous
  • anonymous
yes but it's 1 am where i am and i need 2 to go to bed soon so i guess we can stop here cos i get the gist of it ^-^
anonymous
  • anonymous
but thank u so much for ur help!!! i really appreciate it<3 have a nice night and thank u for all ur amazing help :')
whpalmer4
  • whpalmer4
\[924(2x)^6(-3x^{-2})^6 = 924*2^6x^6*(-3)^6(x^{-2})^6\]\[ = 924*64*729*x^6*x^{-12} \]\[= 43110144x^{-6}\]
anonymous
  • anonymous
(thank u)
whpalmer4
  • whpalmer4
the whole thing is \[\frac{531441}{x^{24}}-\frac{4251528}{x^{21}}+\frac{15588936}{x^{18}}-\frac{34642080}{x^{15}}+4096 x^{12}\]\[+\frac{51963120}{x^{12}}-73728 x^9-\frac{55427328}{x^9}+608256 x^6+\frac{43110144}{x^6}\]\[-3041280 x^3-\frac{24634368}{x^3}+10264320\] :-)
whpalmer4
  • whpalmer4
and now I have met my quota binomial expansion problems for the month :-)

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