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|dw:1441477010872:dw| now square both sides and solve the eqn

\[2\sqrt{4}-2x=2-x\] Rearrange.\[2\sqrt{4}-2=2x-x\]\[\boxed{x=2\sqrt{4}-2}\]

\[\sqrt{4}\]
is this for real?!

No, I think it's four real xD im sorry i had to

You didnt simplify my answer? :(

it's "four real" !!! lol!!

jhannybean, the 2x is also the under the radical. so can we really send it to the other side ?

so your function is \(2\sqrt{4-2x}= 2-x\)?

yeah

um is it (x + 2)(x - 6) = 0 ?

\[(x+2)(x-6) = x^2 -4x-12 \ne x^2+4x-12\]

oh i think i got it
is it (x - 2)( x + 6 ) ?

Yep

then we simplify like this :
x - 2 = 0
x = 2
and
x + 6 = 0
x = - 6
so the answer is x = 2, -6

Mmhmm

omg thank you so much :)

No problem :)

can i ask u one more question ?

Sure

it is similar,
\[\sqrt{5x - x } = 0\]

how do we solve this ?

Square both sides first, right?

Again, we want to isolate x so we need to square both sides. Same method.

wouldnt it just be like 4x = 0 at the end? xD or just plain out 0?

You are correct.

Oooooooo, look at me im doin stuff xD

oh sorry , made a mistake again i meant \[\sqrt{5x} - x = 0 \]

Wouldnt that be the same outcome?

Er... nevermind

@BweadedChicken no, not the same outccome.

oh ok :)