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

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

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

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

tomiko
 2 years ago
Best ResponseYou've already chosen the best response.1for 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.

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

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

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.1good. 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?

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

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

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.1no  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?

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

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

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

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

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

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

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

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

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

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

tomiko
 2 years ago
Best ResponseYou've already chosen the best response.1i 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 ?

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.1ok 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:\[(\sqrt{11}+\sqrt{13})^2=?\]

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

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

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

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

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

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

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.1that 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

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

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

tomiko
 2 years ago
Best ResponseYou've already chosen the best response.1sorry 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.

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

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

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