SOLUTION: Question: sqrt(x)-sqrt(3x-3)=1 My work: sqrt(x)-sqrt(3x-3)=1 (sqrt(x))^2=(1+sqrt(3x-3))^2 x=(1+sqrt(3x-3))(1+sqrt(3x-3)) x=1+sqrt(3x-3)+sqrt(3x-3)+(sqrt(3x-3))^2 x=3x-2+2(

Algebra ->  Radicals -> SOLUTION: Question: sqrt(x)-sqrt(3x-3)=1 My work: sqrt(x)-sqrt(3x-3)=1 (sqrt(x))^2=(1+sqrt(3x-3))^2 x=(1+sqrt(3x-3))(1+sqrt(3x-3)) x=1+sqrt(3x-3)+sqrt(3x-3)+(sqrt(3x-3))^2 x=3x-2+2(      Log On


   

Question 1206414: Question: sqrt(x)-sqrt(3x-3)=1
My work:
sqrt(x)-sqrt(3x-3)=1
(sqrt(x))^2=(1+sqrt(3x-3))^2
x=(1+sqrt(3x-3))(1+sqrt(3x-3))
x=1+sqrt(3x-3)+sqrt(3x-3)+(sqrt(3x-3))^2
x=3x-2+2(sqrt(3x-3))
(-2x+2)/2=2(sqrt(3x-3))/2
(-x)^2=(sqrt(3x-3))^2
x^2=3x-3
0=x^2+3x-3
After this step, I get lost. I know the answer is supposed to be 1 but I am not sure where I went wrong or where to go from the last step I took.
Found 3 solutions by math_tutor2020, ikleyn, MathTherapy:
Answer by math_tutor2020(3817) About Me  (Show Source):
You can put this solution on YOUR website!

I appreciate you showing your work.
The jump from
(-2x+2)/2=2(sqrt(3x-3))/2
to
(-x)^2=(sqrt(3x-3))^2
is incorrect. See lines 6 and 7.

The (-2x+2)/2 should turn into -x+1
The 2s cancel on the right hand side.

This is what you should have for line 7
-x+1 = sqrt(3x-3)

Afterward you would square both sides to get
(-x+1)^2 = 3x-3
I'll let you finish up.
Keep in mind that you'll need to check each possible solution in the original equation to ensure it's an actual solution.

Answer by ikleyn(52781) About Me  (Show Source):
You can put this solution on YOUR website!
.
Question: sqrt(x)-sqrt(3x-3)=1
My work:
sqrt(x)-sqrt(3x-3)=1
(sqrt(x))^2=(1+sqrt(3x-3))^2
x=(1+sqrt(3x-3))(1+sqrt(3x-3))
x=1+sqrt(3x-3)+sqrt(3x-3)+(sqrt(3x-3))^2
x=3x-2+2(sqrt(3x-3))
(-2x+2)/2=2(sqrt(3x-3))/2
(-x)^2=(sqrt(3x-3))^2
x^2=3x-3
0=x^2+3x-3
After this step, I get lost. I know the answer is supposed to be 1 but I am not sure where I went wrong or where to go from the last step I took.
~~~~~~~~~~~~~~~~~~~~~~~~ 

sqrt(x)-sqrt(3x-3)=1

(sqrt(x))^2 = (1+sqrt(3x-3))^2

x = (1+sqrt(3x-3))(1+sqrt(3x-3))

x = 1+sqrt(3x-3)+sqrt(3x-3)+(sqrt(3x-3))^2

x = 3x-2+2(sqrt(3x-3))

(-2x+2) = 2(sqrt(3x-3))

(-2x+2)/2 = 2(sqrt(3x-3))/2     <<<---===  after this line, your writing was incorrect,
                                           so I REPLACED IT. Below is MY writing.

(-x+1)^2 = (sqrt(3x-3))^2

x^2 - 2x + 1 = 3x-3

x^2 - 5x + 4 = 0

(x-4)*(x-1) = 0


So, the two candidates are  x= 4  and  x= 1.
Substitute them into the original equation and make sure
that x= 4 does not work, while x= 1 works.


ANSWER.  The only solution to the original equation is x= 1.

Solved, answered and explained.



Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!
Question: sqrt(x)-sqrt(3x-3)=1

My work: 
sqrt(x)-sqrt(3x-3)=1
(sqrt(x))^2=(1+sqrt(3x-3))^2
x=(1+sqrt(3x-3))(1+sqrt(3x-3))
x=1+sqrt(3x-3)+sqrt(3x-3)+(sqrt(3x-3))^2
x=3x-2+2(sqrt(3x-3))
(-2x+2)/2=2(sqrt(3x-3))/2
(-x)^2=(sqrt(3x-3))^2
x^2=3x-3
0=x^2+3x-3
After this step, I get lost. I know the answer is supposed to be 1 but I am not sure where I went wrong or where to go from the last step I took.

Question: sqrt(x)-sqrt(3x-3)=1
          
                matrix%281%2C3%2C+%28-+x%29%5E2%2C+%22=%22%2C+%28sqrt%283x+-+3%29%29%5E2%29 <===== ERROR IS HERE!!
             matrix%281%2C3%2C+%28-+x+%2B+1%29%5E2%2C+%22=%22%2C+%28sqrt%283x+-+3%29%29%5E2%29 <===== s/b THIS, instead!
     
     (x - 1)  (x - 4) = 0
x - 1 = 0   |   x - 4 = 0 ---- Equating each factor to 0
    x = 1   |       x = 4

                        C H E C K
      x = 1                               x = 4
                  
             1 = 1 (TRUE)                - 1 = 1 (FALSE)
          x = 1 is the ONLY SOLUTION     x = 4 is an EXTRANEOUS SOLUTION