SOLUTION: Find solution to : [sqrt](2k-1) - [sqrt](k-1) = 1 So far I have added [sqrt](k-1) to both sides of equation in order to isolate one radical. Now I have to square both sid

Algebra ->  Radicals -> SOLUTION: Find solution to : [sqrt](2k-1) - [sqrt](k-1) = 1 So far I have added [sqrt](k-1) to both sides of equation in order to isolate one radical. Now I have to square both sid      Log On


   

Question 364929: Find solution to :
[sqrt](2k-1) - [sqrt](k-1) = 1
So far I have added [sqrt](k-1) to both sides of equation in order to isolate one radical.
Now I have to square both sides
([sqrt](2k-1))squared = ([sqrt](k-1) +1)squared
2k-1= ([sqrt](k-1)+1))*([sqrt](k-1)+1))
Somewhere later in my problem I think I am not simplifying correctly. I ended up with :
(k-1)/2 = [sqrt]k-1
so then I was back to squaring both sides of my equation

Found 2 solutions by amoresroy, robertb:
Answer by amoresroy(361) About Me  (Show Source):
You can put this solution on YOUR website!
(k - 1) / 2 = [sqrt](k - 1)
k -1 = 2 [sqrt](k - 1)
Square both sides of the equation
-----------------------------------
k^2 - 2k + 1 = 4 (k - 1)
k^2 - 2k + 1 = 4k - 4
k^2 - 6k + 5 = 0
(k - 1) (k - 5) = 0
---------------------------
k - 1 = 0
k = 1
-----------------
k - 5 =0
k = 5
------------------------
So the values of k are 1 and 5.

Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
sqrt%282k-1%29+-+sqrt%28k-1%29+=+1
Transpose to get sqrt%282k-1%29+=+sqrt%28k-1%29+%2B+1.
Square both sides: 2k-1+=+1%2B2sqrt%28k-1%29+%2B+k-1.
Simplify: k-1+=+2sqrt%28k-1%29.
Square both sides again: k%5E2-2k%2B1+=+4%28k-1%29,
k%5E2-2k%2B1+=+4k-4%29,
or k%5E2+-+6k%2B5+=+0,
then (k-5)(k-1) = 0, or k = 5 or k = 1.
Both satisfy the original equation, so the solution set is {1,5}.