Radical Equations - resolved

X

Xearf_987

New member
Joined
Feb 14, 2006
Messages
22
The problem: sqrt(3x) + sqrt(2x-1) = 5 / sqrt(2x-1)

I'm suppose to solve for x. I'm not sure if I started this problem correctly or not, but here's what I got.

My attempt:

[I multiplied both sides of the equation by sqrt(2x-1) to eliminate the denomonater on the right side of the equation]

sqrt(6x-1) + sqrt(4x^2-4x+1) = 5

[then I isolated one radical]

sqrt(6x-1) = 5 - sqrt(4x^2-4x+1)

[I squared both sides]

6x - 1 = 25 - 10sqrt(4x^2-4x+1) + 4x^2 - 4x + 1

[combined like terms]

10sqrt(4x^2-4x+1) = 4x^2 - 10x +27

This is about where I decided I must have done something wrong. Am I on the right track or am I not even close... or what?
 
i don't think you can add the sqrt if it's not the same inside because they are different numbers

see, 3x does not equal 2x-1, therefore, your first step is wrong if i'm right, hope this helps
 
HINT: square the both sides first in the equation [this thing, ^2] and keep solving is and you'll get a quadric formula, hope this helps
 
First of all, I didn't add =/ Can you quote what you're talking about? I know that you can't add square roots of different values. I know for a fact I didn't do that (unless I'm absolutely insane and made the careless mistake without realizing it). And I did square both sides of the equation. You have to isolate a radical before you can do that though, which I did. I hope I'm understanding you correctly >< Thanks for your help though! However, I'm still not clear on what exactly I didn't do right.
 
Right idea but
sqrt(3x)*sqrt(2x-1) =
sqrt(6x²-3x)
Also note
sqrt(2x-1)*sqrt(2x-1) = 2x-1
(as long as it is positive)
Try it from there.
 
sorry, not so good in explaining with details, i'll try better next time ^^
 
oh...also, i was doing in a different way too...oops
heh, guess i went a bit ahead, sorry
i didn't multiply first...but solved the sqrt first
 
You must log in or register to reply here.
Top