Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

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

See more answers at
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly


Get your free account and access expert answers to this and thousands of other questions

@tomiko are you aware of this factorisation?\[a^2-b^2=(a+b)(a-b)\]
yes. but i don't know how to apply this to work out the answer. please help me
look at the first term:\[\sqrt{12}-\sqrt{11}\]if you multiply this by \(\sqrt{12}+\sqrt{11}\) what will you get?

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

you get 1
correct. nor similarly multiply the second expression with \(\sqrt{13}+\sqrt{12}\) - what do you get?
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.
good. now, out these two terms, which one is bigger:\[1. \sqrt{12}+\sqrt{11}\]\[2.\sqrt{12}+\sqrt{13}\]
\[\sqrt{12} + \sqrt{13}\] is bigger
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?
I meant to say: you know (2) is bigger than (1)
yes. 2 is bigger than one so that makes \[\sqrt{13} - \sqrt{12}\] bigger?? :D
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?
in that case x is bigger
imagine you had:\[x*4=1\]\[y*8=1\]which one would you say is bigger, x or y?
yes, x is bigger, therefore you know which one is bigger in your original question
yes i know \[\sqrt{13} - \sqrt{12}\] is bigger.
look back carefully through what was derived and re-think your answer.
remember that \(\sqrt{13} - \sqrt{12}\) was multiplied by the bigger number \(\sqrt{13} + \sqrt{12}\) to get 1.
yes!! i get it now. had to look at my solutions again using the different of 2 squares.
we showed that:\[(\sqrt{12}-\sqrt{11})(\sqrt{12}+\sqrt{11})=1\]\[(\sqrt{13}-\sqrt{12})(\sqrt{13}+\sqrt{12})=1\]and\[\sqrt{13}+\sqrt{12}\gt\sqrt{12}+\sqrt{11}\]
that makes \[\sqrt{12} - \sqrt{11}\] greater!
perfect! well done! :)
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 ?
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:\[(\sqrt{11}+\sqrt{13})^2=?\]
do you understand?
yeap...! i'm trying to do it on paper here :D
ok :)
and this shows \[\sqrt{12} - \sqrt{11}\] is bigger. you can see my working attached as an image
1 Attachment
you expansion of \((\sqrt{11}+\sqrt{13})^2\) is not correct. remember that:\[(a+b)^2=a^2+2ab+b^2\]
so if i correct it. i get \[48 ? 24 + 2\times \sqrt{13}\times \sqrt{11}\]
that is correct so far. now take the 24 to the left hand side and then divide both sides by 2
also remember that:\[\sqrt{a}\times\sqrt{b}=\sqrt{ab}\]
now i get (attached)
1 Attachment
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
now the final answer!! :d
great! I'm glad you finally got there :)
1 Attachment
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.
no problem at all my friend - the main thing here is you obviously love to learn! :)
challenge** thank you very much once again. now any question like this that comes my way i will murder it!!
he he - I love your "ambition" - keep up the good work! :)

Not the answer you are looking for?

Search for more explanations.

Ask your own question