how do i work out which number is greater?
(sqrt12 - sqrt11), (sqrt13 - sqrt12)

- anonymous

how do i work out which number is greater?
(sqrt12 - sqrt11), (sqrt13 - sqrt12)

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- schrodinger

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- asnaseer

@tomiko are you aware of this factorisation?\[a^2-b^2=(a+b)(a-b)\]

- anonymous

yes. but i don't know how to apply this to work out the answer. please help me

- asnaseer

look at the first term:\[\sqrt{12}-\sqrt{11}\]if you multiply this by \(\sqrt{12}+\sqrt{11}\) what will you get?

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- anonymous

you get 1

- asnaseer

correct. nor similarly multiply the second expression with \(\sqrt{13}+\sqrt{12}\) - what do you get?

- asnaseer

*now

- anonymous

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.

- asnaseer

good. now, out these two terms, which one is bigger:\[1. \sqrt{12}+\sqrt{11}\]\[2.\sqrt{12}+\sqrt{13}\]

- anonymous

\[\sqrt{12} + \sqrt{13}\] is bigger

- asnaseer

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?

- asnaseer

I meant to say: you know (2) is bigger than (1)

- anonymous

yes. 2 is bigger than one so that makes \[\sqrt{13} - \sqrt{12}\] bigger?? :D

- asnaseer

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?

- anonymous

in that case x is bigger

- asnaseer

imagine you had:\[x*4=1\]\[y*8=1\]which one would you say is bigger, x or y?

- anonymous

x

- asnaseer

yes, x is bigger, therefore you know which one is bigger in your original question

- anonymous

yes i know \[\sqrt{13} - \sqrt{12}\] is bigger.

- asnaseer

look back carefully through what was derived and re-think your answer.

- asnaseer

remember that \(\sqrt{13} - \sqrt{12}\) was multiplied by the bigger number \(\sqrt{13} + \sqrt{12}\) to get 1.

- anonymous

yes!! i get it now. had to look at my solutions again using the different of 2 squares.

- asnaseer

we showed that:\[(\sqrt{12}-\sqrt{11})(\sqrt{12}+\sqrt{11})=1\]\[(\sqrt{13}-\sqrt{12})(\sqrt{13}+\sqrt{12})=1\]and\[\sqrt{13}+\sqrt{12}\gt\sqrt{12}+\sqrt{11}\]

- anonymous

that makes \[\sqrt{12} - \sqrt{11}\] greater!

- asnaseer

perfect! well done! :)

- anonymous

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 ?

- asnaseer

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:\[(\sqrt{11}+\sqrt{13})^2=?\]

- asnaseer

do you understand?

- anonymous

yeap...! i'm trying to do it on paper here :D

- asnaseer

ok :)

- anonymous

and this shows \[\sqrt{12} - \sqrt{11}\] is bigger. you can see my working attached as an image

##### 1 Attachment

- asnaseer

you expansion of \((\sqrt{11}+\sqrt{13})^2\) is not correct. remember that:\[(a+b)^2=a^2+2ab+b^2\]

- anonymous

so if i correct it. i get
\[48 ? 24 + 2\times \sqrt{13}\times \sqrt{11}\]

- asnaseer

that is correct so far. now take the 24 to the left hand side and then divide both sides by 2

- asnaseer

also remember that:\[\sqrt{a}\times\sqrt{b}=\sqrt{ab}\]

- anonymous

now i get (attached)

##### 1 Attachment

- asnaseer

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

- anonymous

now the final answer!! :d

- asnaseer

great! I'm glad you finally got there :)

- anonymous

attached

##### 1 Attachment

- asnaseer

perfect!

- anonymous

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.

- asnaseer

no problem at all my friend - the main thing here is you obviously love to learn! :)

- anonymous

challenge**
thank you very much once again. now any question like this that comes my way i will murder it!!

- asnaseer

he he - I love your "ambition" - keep up the good work! :)

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