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Give me a second while I type out the expression. Thank you

\(\huge\frac{\sqrt{10}}{\sqrt{5}-2}\)

do you know the root of ten?

\(\huge\sqrt{10} = 3.16227766...\)

do you want to see if you can do anything by multiplying it by a conjugate

a conjugate...so like the opposite? do you mean multiplying by \(\huge\sqrt{5}+2\) ??

o-o

|dw:1440462611069:dw|

root of 30 is irrational

is that denominator \(\huge \sqrt{5}-2~ or~ \sqrt{5-2}\)

wait lemme check mein textbook

it is definitely \(\huge\sqrt{5}-2\)

|dw:1440462940673:dw|

oh right! inverses :S

let me show you another example
\((\sqrt{2} +3) \times (\sqrt{2}-3) = 2-9 = 7 \)

oops -7 not 7 haha

right, so there wouldn't be a root over the two any more

I recall that rule: \[\sqrt{x^{2}}\] = x

i see i see.

you could do|dw:1440463196135:dw|

how would you multiply on the numerator though?

ur answer is 2root10 plus 5root2

@nincompoop ur making it way harder than it is

heres the answer \[2\sqrt{10}+5\sqrt{2}\]

Oh I see. so you basically simplified \(\sqrt{50}\) Down to \(5\sqrt{2}\). I understand that

yeeeeee
now try your own problem

as for the \(2\sqrt{10}\), that's just 2 x \(\sqrt{10}\). Got it!

well you make everything way harder than it already is

*nin
oops my bad

i'd just help the person that let him get an F

*than

*then
and that's not a good philosophy

enough bs in the thread. move on to the next question

scru dis