Question

Solve for x. √2x+3=√6x-1 1 2 3 √6

Answer 1

is x under the square root??

Answer 2

yes.

Answer 3

+3 and -1 is under square root?

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

@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.

Answer 8

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

Answer 9

\[\sqrt{6x}-\sqrt{2x}=4\]

\[\sqrt{3}\cdot \sqrt{2x}-\sqrt{2x}=4\]

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

Answer 10

ok now i solve it! please wait!!!

Answer 11

okay.

Answer 12

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

Answer 13

i'm waiting.

Answer 14

I'm waiting too :D

Answer 15

lol ;)

Answer 16

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.

Answer 17

the second one was the question ...

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

√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!!!

Answer 22

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

2x+3=6x-1

2x-6x=-1-3

-3x=-4

\[x=\frac{-4}{-3} \longrightarrow x=\frac{4}{3}\]

Answer 23

@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...

Answer 24

I'm drawing right now...

Answer 25

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