Solve for x. √2x+3=√6x-1 1 2 3 √6

- anonymous

Solve for x. √2x+3=√6x-1 1 2 3 √6

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- anonymous

is x under the square root??

- anonymous

yes.

- anonymous

+3 and -1 is under square root?

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## More answers

- anonymous

I can't do it ... it seems the X cancels ... maybe I'm doing a mistake somewhere. !

- anonymous

\[\sqrt{6x} - \sqrt{2x} = \sqrt{4x} \rightarrow 3 + 1 = 4 \rightarrow \sqrt{4x} = 4\]
now square both sides
\[\sqrt{4x}^{2} = 4^{2} \rightarrow 4x = 16 \rightarrow x = 16 \div 4 \rightarrow x = 4\]

- anonymous

@JoshDavoll your answer is wrong!!!
because √6x - √2x != √4x

- anonymous

@alirezamemarian - yes
@Kreshnik - the numbers at the end are the choices. they clumped together for some reason. I tried to change them but it wouldn't let me.
@JoshDavoll - that isn't one of the choices.

- anonymous

\[\sqrt{a} - \sqrt{b} = \sqrt{a - b}\]

- anonymous

\[\sqrt{6x}-\sqrt{2x}=4\]
\[\sqrt{3}\cdot \sqrt{2x}-\sqrt{2x}=4\]
substitute \[\sqrt{2x}=a \]
and you'll have \[3a-a=4 \Longrightarrow 2a=4 \Longrightarrow a=2\]
substitute back
\[\sqrt{2x}=2\] square both sides
2x=4
x=2

- anonymous

ok now i solve it! please wait!!!

- anonymous

okay.

- anonymous

wait till @alirezamemarian finishes his work... it would be unfair if you leave :$ ... wait on :D

- anonymous

i'm waiting.

- anonymous

I'm waiting too :D

- anonymous

lol ;)

- anonymous

is the question \[\sqrt{2x + 3} = \sqrt{6x - 1} \] or \[\sqrt{2x} + 3 = \sqrt{6x} - 1\] ?
My answer wasn't wrong i just answered for the 2nd question. Sorry.

- anonymous

the second one was the question ...

- anonymous

@JoshDavoll - the first one is correct. I just don't know how to put that long bar on the top of the square root.

- anonymous

I made a mistake... it seems I turned \[\sqrt{3}] into 3 LOL .... then my answer is wrong... forget IT I'll try again :S

- anonymous

okay. Thanks everyone who is trying to help me! I really appreciate it! :)

- anonymous

√6x = √2x + 4
square both side
6x= 2x + 16 +8√2x
4x-16=8√2x
square both side
(4x-16)^2=64*2x
\[16x ^{2}-256-128x=128x\]
\[\div16\]
\[x ^{2}-16-16x=0\]
\[\Delta=24\]
\[x=(-b \pm \Delta)/a\]
x1=20
x2=-4
that x2 is Unacceptable
finally x=20!!!

- anonymous

\[\sqrt{2x+3}=\sqrt{6x-1}\] square both sides and you have:
2x+3=6x-1
2x-6x=-1-3
-3x=-4
\[x=\frac{-4}{-3} \longrightarrow x=\frac{4}{3}\]

- anonymous

@emdrais take the pencil and DRAW IT because no one got the same task you're asking for... I'm confused now... write it well again please .. WITH PENCIL LIKE THIS...

- anonymous

|dw:1333177883692:dw|

- anonymous

|dw:1333177956843:dw|I'm drawing right now...

- anonymous

then \[x=\frac43\] check out my previous post... that's correct !

- anonymous

|dw:1333178066093:dw|

- anonymous

FINALLY x=1 HERE"S THE MISTAKE I MADE!! \[\LARGE 2x-6x\neq -3x \longrightarrow 2x-6x=-4x\] just put that on and you're ready to go...

- anonymous

say it first!!!
you loss my time!!! :D
2x+3=6x-1
4=4x
x=1!!!

- anonymous

@alirezamemarian you're right... that's correct !

- anonymous

@Kreshnik - It's on an online quiz
@alirezamemarian - is 1 correct?

- anonymous

yes 100% x=1

- anonymous

yes x=1 100% ;)

- anonymous

\[\sqrt{2x + 3} = \sqrt{6x - 1}\]
this is the same as:
\[\sqrt{2x} + \sqrt{3} = \sqrt{6x} - \sqrt{1}\]
subtract \[\sqrt{2x}\] from both sides, they cancel out on the right.
\[\sqrt{4x} - \sqrt{1} = \sqrt{3}\]
add \[\sqrt{1}\] to both sides:
\[\sqrt{4x} = \sqrt{4}\]
now square both sides and solve for x. x = 1 (sorry took ages to type)

- anonymous

lol oh well, we got there in the end

- anonymous

@JoshDavoll .... even if your answer is correct you made a mistake.... \[\LARGE \sqrt{2x+3}\neq \sqrt{2x}+\sqrt3 \]

- anonymous

yes it is! :)

- anonymous

that's how my answer is correct ;)

- anonymous

\[\LARGE \sqrt{2x\cdot 3}=\sqrt{2x}\cdot \sqrt3\] but not with plus or minus !

- anonymous

you can distribute the square root sign anywhere you like :)

- anonymous

Thanks everyone for the help!!

- anonymous

and you're totally wrong... \[\sqrt{6x}-\sqrt{2x}\neq \sqrt3\] I don't get you at all...

- anonymous

where did i say that???

- anonymous

i never said it was equal to \[\sqrt{3}\] are you wearing your glasses?

- anonymous

.... if you didn't that means your post is worthless... can you explain me what you did in your previous post... DRAW IT if you have a problem with latex... let's see if my glasses are right??

- anonymous

i think you have a problem with viewing ! just read my post and tell me where the problem is

- anonymous

Guys don't fight!! I'm going to close the thread now, but thank you all for your help!! ;)

- anonymous

i subtracted square of 2x from both sides from 2x and 4x, the two 2x's cancelled out and i was left with square of 4x - square of 1 = square of 3 then i flipped the equation so the square of 4x was in front. Does that make sense to you??

- anonymous

@JoshDavoll http://i1064.photobucket.com/albums/u368/Kreshnik-joker/SQRT.jpg THIS IS MY REPLY ... sorry for being late. BYE !

- anonymous

then why was i correct, buddy? there are more than one way to do things @Kreshnik

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