Question

Help

Answer 1

Have do I come from this
\[\sqrt{n^2+nx}-1\]
to this
\[\Large \frac{\sqrt{1+\frac{x}n}-1}{\frac1n}\]

Answer 2

\[\sqrt{n ^{2}+nx}-1\]=\[\frac{ \sqrt{1+\frac{ x }{ n }}-\frac{ 1 }{ n }}{ \frac{ 1 }{ n } }\]

Answer 3

i don't see any way, because those 2 aren't equal

Answer 4

what souvik did is correct

Answer 5

you want to know how to get there ?

Answer 6

arrh I mean \[\sqrt{n^2+nx}-n\]

Answer 7

then what you wrote is correct

Answer 8

then you are right!

Answer 9

can you factor out n^2 from n^2+nx ?

Answer 10

no ?
then, can you factor out n from n^2+nx first?

Answer 11

\[\sqrt{n*(n+x)}-n\]

Answer 12

no, thats incorrect,
factor out 'n' from n+x

Answer 13

actually \(\sqrt{n*(n+x)}=\sqrt n \sqrt{n+x}\)

Answer 14

I cant spot the solution.. :(

Answer 15

you got this, right ?? \(\sqrt{n*(n+x)}=\sqrt n \sqrt{n+x}\)

now n+x = n*1 + x/n * n = n (1+x/n)
clear rtill here ?doubts ?

Answer 16

I can follow you until here:
\[\sqrt{n^2+nx}-n=\sqrt{n*(n+x)}-n=\sqrt{n} \sqrt{n+x}-n\]

Answer 17

ok, can you distribute ?
a (b+c) = ab+ac
what about \(n \times (1+x/n) = ... ?\)

Answer 18

\[=n+x\]

Answer 19

so, i just replaced n+x by that quantity, by 'undistributing' or 'factoring' out the n

Answer 20

Yes so we have:
\[\sqrt{n}\sqrt{n*(1+\frac{ x }{ n })}-n\]

Answer 21

\(n+x=(n\times 1+(x/n)\times n)=n(1+x/n)\)

Answer 22

yeah,
\(\large \sqrt{n}\sqrt{n*(1+\frac{ x }{ n })}-n=\sqrt{n}\times \sqrt n\times \sqrt{(1+\frac{ x }{ n })}-n\)

and \(\sqrt n \sqrt n=n\)

Answer 23

ok ?

Answer 24

\[\sqrt{n}\sqrt{n}\sqrt{1+\frac{ x }{ n }}-n=\frac{ \frac{ n*\sqrt{1+\frac{ x }{ n }}-n }{ n } }{ \frac{ 1 }{ n } }=\frac{ \sqrt{1+\frac{ x }{ n }}-1 }{ \frac{ 1 }{ n } }\]

Answer 25

you got it! :)

Answer 26

Thank you.

Answer 27

welcome ^_^