anonymous
  • anonymous
Solve for x. √2x+3=√6x-1 1 2 3 √6
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
is x under the square root??
anonymous
  • anonymous
yes.
anonymous
  • anonymous
+3 and -1 is under square root?

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anonymous
  • anonymous
I can't do it ... it seems the X cancels ... maybe I'm doing a mistake somewhere. !
anonymous
  • 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
  • anonymous
@JoshDavoll your answer is wrong!!! because √6x - √2x != √4x
anonymous
  • 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
  • anonymous
\[\sqrt{a} - \sqrt{b} = \sqrt{a - b}\]
anonymous
  • 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
  • anonymous
ok now i solve it! please wait!!!
anonymous
  • anonymous
okay.
anonymous
  • anonymous
wait till @alirezamemarian finishes his work... it would be unfair if you leave :$ ... wait on :D
anonymous
  • anonymous
i'm waiting.
anonymous
  • anonymous
I'm waiting too :D
anonymous
  • anonymous
lol ;)
anonymous
  • 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
  • anonymous
the second one was the question ...
anonymous
  • 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
  • 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
  • anonymous
okay. Thanks everyone who is trying to help me! I really appreciate it! :)
anonymous
  • 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
  • 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
  • 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
  • anonymous
|dw:1333177883692:dw|
anonymous
  • anonymous
|dw:1333177956843:dw|I'm drawing right now...
anonymous
  • anonymous
then \[x=\frac43\] check out my previous post... that's correct !
anonymous
  • anonymous
|dw:1333178066093:dw|
anonymous
  • 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
  • anonymous
say it first!!! you loss my time!!! :D 2x+3=6x-1 4=4x x=1!!!
anonymous
  • anonymous
@alirezamemarian you're right... that's correct !
anonymous
  • anonymous
@Kreshnik - It's on an online quiz @alirezamemarian - is 1 correct?
anonymous
  • anonymous
yes 100% x=1
anonymous
  • anonymous
yes x=1 100% ;)
anonymous
  • 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
  • anonymous
lol oh well, we got there in the end
anonymous
  • anonymous
@JoshDavoll .... even if your answer is correct you made a mistake.... \[\LARGE \sqrt{2x+3}\neq \sqrt{2x}+\sqrt3 \]
anonymous
  • anonymous
yes it is! :)
anonymous
  • anonymous
that's how my answer is correct ;)
anonymous
  • anonymous
\[\LARGE \sqrt{2x\cdot 3}=\sqrt{2x}\cdot \sqrt3\] but not with plus or minus !
anonymous
  • anonymous
you can distribute the square root sign anywhere you like :)
anonymous
  • anonymous
Thanks everyone for the help!!
anonymous
  • anonymous
and you're totally wrong... \[\sqrt{6x}-\sqrt{2x}\neq \sqrt3\] I don't get you at all...
anonymous
  • anonymous
where did i say that???
anonymous
  • anonymous
i never said it was equal to \[\sqrt{3}\] are you wearing your glasses?
anonymous
  • 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
  • anonymous
i think you have a problem with viewing ! just read my post and tell me where the problem is
anonymous
  • anonymous
Guys don't fight!! I'm going to close the thread now, but thank you all for your help!! ;)
anonymous
  • 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
  • anonymous
@JoshDavoll http://i1064.photobucket.com/albums/u368/Kreshnik-joker/SQRT.jpg THIS IS MY REPLY ... sorry for being late. BYE !
anonymous
  • anonymous
then why was i correct, buddy? there are more than one way to do things @Kreshnik

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