SOLUTION: I am having a bit of an issue solving sqrt(3x+1)-sqrt(x-1)=2 or √3x+1-√x-1 = 2
This is what I have done so far:
√3x+1 = √x-1+2
=> (√3x+1
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Question 1068538: I am having a bit of an issue solving sqrt(3x+1)-sqrt(x-1)=2 or √3x+1-√x-1 = 2
This is what I have done so far:
√3x+1 = √x-1+2
=> (√3x+1)2 = (√x-1+2)2
=> 3x+1 = x-1+4+4√x-1
=> -4√x-1 = -3x-1+x-1+4 or is it 4 √x+1 = -3x-1+x-1+4?
=> 4√x-1 = 2x-2 *This is where I get lost. I have seen this as -2x+2 as well. I am really confused at this point and need to get to the next steps.
Thank you for your help in advance.
Found 3 solutions by Fombitz, Boreal, Alan3354:
Answer by Fombitz(32388) (Show Source): You can put this solution on YOUR website!
is correct.
Keep going.
Now square both sides again.
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
√3x+1 = √x-1+2
=> (√3x+1)^2 = 2+ sqrt (x-1)
=> 3x+1 = 4 +4 sqrt(x-1)+x-1. I like to bring the smaller x to the side with the larger and keep it positive but one doesn't have to do that.
=2x-2=4 sqrt(x-1).
***square both sides***remember to square the 4, too.
4x^2-8x+4=16(x-1)=16x-16
4x^2-24x+20=0
x^2-6x+5=0
(x-5)(x-1)=0
x=5 and 1
sqrt(16)-sqrt(4)=4-2=2 checks
sart(4)-sqrt(0)=2 also checks.
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
I am having a bit of an issue solving sqrt(3x+1)-sqrt(x-1)=2 or √3x+1-√x-1 = 2
This is what I have done so far:
√3x+1 = √x-1+2
=> (√3x+1)2 = (√x-1+2)2
=> 3x+1 = x-1+4+4√x-1
=> -4√x-1 = -3x-1+x-1+4 or is it 4 √x+1 = -3x-1+x-1+4?
-4sqrt(x-1) = -3x-1+x-1+4 works.
-4sqrt(x-1) = -2x+2
-----
16(x-1) = 4x^2 - 8x + 4
4(x-1) = x^2 - 2x + 1
4x-4 = x^2 - 2x + 1
x^2 - 6x + 5 = 0
(x-1)*(x-5) = 0
x = 1, 5
=======================
=> 4√x-1 = 2x-2 *This is where I get lost. I have seen this as -2x+2 as well.
It doesn't matter. Comes out the same after it's squared.
---
I am really confused at this point and need to get to the next steps.
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