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Mathematics 11 Online
OpenStudy (anonymous):

help how do u solve sqrt(x)+sqrt(2x)=1

OpenStudy (anonymous):

3-2sqrt(2)

OpenStudy (anonymous):

is it possible to explain or show me how you got this

OpenStudy (anonymous):

yes, first square both sides

OpenStudy (anonymous):

add them together you get (3+2sqrt(2))x=1

OpenStudy (anonymous):

this is tricky, you realize that if you multiply by the complement, you get 9-8 which equals 1

OpenStudy (anonymous):

oh ok i understand it now i know ho to do the ones with one sqrt but wasnt sure how to do it with two sqrt's thanks

OpenStudy (anonymous):

sometimes it takes just a little flash of insight

OpenStudy (anonymous):

wait i just got lost were did you get the 3 from

OpenStudy (radar):

\[(\sqrt{x}+\sqrt{2x})^{2}\]

OpenStudy (radar):

that should result in: 3x plusminus (2x)2^1/2. But he omitted squaring the other side getting a 1

OpenStudy (radar):

and the equating that ugly thing to 1 and solving, so far I haven't seen a solution.

OpenStudy (anonymous):

well the solution is 3-2\[\sqrt{2}\]

OpenStudy (anonymous):

i mean t 3-2sqrt(2)

OpenStudy (anonymous):

but i dont know how he got that i got lost after i looked back again

OpenStudy (anonymous):

see, the thing is (3+2sqrt(2))(3-2sqrt(2))=(3)^2-(2sqrt(2))^2

OpenStudy (anonymous):

which is 9-8=1

OpenStudy (anonymous):

the COMPLEMENT of (3+2sqrt(2)) is (3-2sqrt(2)) b/c that is like squaring both elements.

OpenStudy (radar):

So your using + root, a - root. Is that a fair statement?

OpenStudy (radar):

It does check out, in a baffling sort of way.

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