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tomiko
Group Title
how do i work out which number is greater?
(sqrt12  sqrt11), (sqrt13  sqrt12)
 2 years ago
 2 years ago
tomiko Group Title
how do i work out which number is greater? (sqrt12  sqrt11), (sqrt13  sqrt12)
 2 years ago
 2 years ago

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asnaseer Group TitleBest ResponseYou've already chosen the best response.1
@tomiko are you aware of this factorisation?\[a^2b^2=(a+b)(ab)\]
 2 years ago

tomiko Group TitleBest ResponseYou've already chosen the best response.1
yes. but i don't know how to apply this to work out the answer. please help me
 2 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.1
look at the first term:\[\sqrt{12}\sqrt{11}\]if you multiply this by \(\sqrt{12}+\sqrt{11}\) what will you get?
 2 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.1
correct. nor similarly multiply the second expression with \(\sqrt{13}+\sqrt{12}\)  what do you get?
 2 years ago

tomiko Group TitleBest ResponseYou've already chosen the best response.1
for that too i get 1. but one is definately greater. cause when i use the calculate i get 2 different values, making one of the pairs greater than the other.
 2 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.1
good. now, out these two terms, which one is bigger:\[1. \sqrt{12}+\sqrt{11}\]\[2.\sqrt{12}+\sqrt{13}\]
 2 years ago

tomiko Group TitleBest ResponseYou've already chosen the best response.1
\[\sqrt{12} + \sqrt{13}\] is bigger
 2 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.1
good. so you now know that: a) x times (1) = 1 b) y times (2) = 1 and you know (2) is bigger than one. therefore what can you conclude about x and y?
 2 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.1
I meant to say: you know (2) is bigger than (1)
 2 years ago

tomiko Group TitleBest ResponseYou've already chosen the best response.1
yes. 2 is bigger than one so that makes \[\sqrt{13}  \sqrt{12}\] bigger?? :D
 2 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.1
no  in terms of x and y that I listed, x is being multiplied by a smaller number than y is being multiplied by, but both end up as 1. therefore is x or y bigger?
 2 years ago

tomiko Group TitleBest ResponseYou've already chosen the best response.1
in that case x is bigger
 2 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.1
imagine you had:\[x*4=1\]\[y*8=1\]which one would you say is bigger, x or y?
 2 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.1
yes, x is bigger, therefore you know which one is bigger in your original question
 2 years ago

tomiko Group TitleBest ResponseYou've already chosen the best response.1
yes i know \[\sqrt{13}  \sqrt{12}\] is bigger.
 2 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.1
look back carefully through what was derived and rethink your answer.
 2 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.1
remember that \(\sqrt{13}  \sqrt{12}\) was multiplied by the bigger number \(\sqrt{13} + \sqrt{12}\) to get 1.
 2 years ago

tomiko Group TitleBest ResponseYou've already chosen the best response.1
yes!! i get it now. had to look at my solutions again using the different of 2 squares.
 2 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.1
we showed that:\[(\sqrt{12}\sqrt{11})(\sqrt{12}+\sqrt{11})=1\]\[(\sqrt{13}\sqrt{12})(\sqrt{13}+\sqrt{12})=1\]and\[\sqrt{13}+\sqrt{12}\gt\sqrt{12}+\sqrt{11}\]
 2 years ago

tomiko Group TitleBest ResponseYou've already chosen the best response.1
that makes \[\sqrt{12}  \sqrt{11}\] greater!
 2 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.1
perfect! well done! :)
 2 years ago

tomiko Group TitleBest ResponseYou've already chosen the best response.1
i undertand now. thank you very much! but you have any idea about this approach? Purely algebra ? Place ? i/o =, > or <. sqrt(12)sqrt(11) ? sqrt(13)sqrt(12) 2sqrt(12) ? sqrt(11)+sqrt(13) 48 ? ... Can you continue ?
 2 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.1
ok what this approach is doing is bringing together the two square roots of 12 and and then squaring both sides. so:\[(2\sqrt{12})^2=4*12=48\]now you just need to square the right hand side:\[(\sqrt{11}+\sqrt{13})^2=?\]
 2 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.1
do you understand?
 2 years ago

tomiko Group TitleBest ResponseYou've already chosen the best response.1
yeap...! i'm trying to do it on paper here :D
 2 years ago

tomiko Group TitleBest ResponseYou've already chosen the best response.1
and this shows \[\sqrt{12}  \sqrt{11}\] is bigger. you can see my working attached as an image
 2 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.1
you expansion of \((\sqrt{11}+\sqrt{13})^2\) is not correct. remember that:\[(a+b)^2=a^2+2ab+b^2\]
 2 years ago

tomiko Group TitleBest ResponseYou've already chosen the best response.1
so if i correct it. i get \[48 ? 24 + 2\times \sqrt{13}\times \sqrt{11}\]
 2 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.1
that is correct so far. now take the 24 to the left hand side and then divide both sides by 2
 2 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.1
also remember that:\[\sqrt{a}\times\sqrt{b}=\sqrt{ab}\]
 2 years ago

tomiko Group TitleBest ResponseYou've already chosen the best response.1
now i get (attached)
 2 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.1
that is correct  just one step left now  square both sides BTW: It is better to save your images as .png format if you can  they are smaller than .jpg files
 2 years ago

tomiko Group TitleBest ResponseYou've already chosen the best response.1
now the final answer!! :d
 2 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.1
great! I'm glad you finally got there :)
 2 years ago

tomiko Group TitleBest ResponseYou've already chosen the best response.1
sorry for all the bother. thanks for the help. i finished high school 6 years ago and didn't do any math at all and now just started uni. it's going to be a little change for me. i'm doing computer science now. lots of math in my class!! but i love subject.
 2 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.1
no problem at all my friend  the main thing here is you obviously love to learn! :)
 2 years ago

tomiko Group TitleBest ResponseYou've already chosen the best response.1
challenge** thank you very much once again. now any question like this that comes my way i will murder it!!
 2 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.1
he he  I love your "ambition"  keep up the good work! :)
 2 years ago
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