SOLUTION: I need to solve this equation for x: sqrt(2x-1) - sqrt(x-1) = 1 This is what I have so far: sqrt(2x-1)=sqrt(x-1)+1 2x+1=((sqrt(x-1) +1))((sqrt(x-1)+1) 2x-1=x-1+2sqrt(x-1)+

Algebra ->  Radicals -> SOLUTION: I need to solve this equation for x: sqrt(2x-1) - sqrt(x-1) = 1 This is what I have so far: sqrt(2x-1)=sqrt(x-1)+1 2x+1=((sqrt(x-1) +1))((sqrt(x-1)+1) 2x-1=x-1+2sqrt(x-1)+      Log On


   

Question 801875: I need to solve this equation for x:
sqrt(2x-1) - sqrt(x-1) = 1

This is what I have so far:
sqrt(2x-1)=sqrt(x-1)+1
2x+1=((sqrt(x-1) +1))((sqrt(x-1)+1)
2x-1=x-1+2sqrt(x-1)+1
(x+1)^2=(2sqrt(x-1))^2
(x+1)(x+1)=4(x-1)
x^2+2x+1=4x-4
x^2-2x+5=0
I know something is wrong because it doesn't factor.
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


This is what I have so far:
sqrt(2x-1)=sqrt(x-1)+1
2x+1=((sqrt(x-1) +1))((sqrt(x-1)+1)
2x-1=x-1+2sqrt(x-1)+1
Sign error right here:
(x+1)^2=(2sqrt(x-1))^2

(x+1)(x+1)=4(x-1)
x^2+2x+1=4x-4
x^2-2x+5=0













That should factor quite tidily. However, don't depend on everything to factor neatly. Some quadratics have irrational zeros and others have complex zeros. Also, whenever you square both sides of an equation in the process of solving it, you must consider the possibility that you have introduced extraneous roots. CHECK ALL ANSWERS!

John

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