Radical equation question

I keep ending up with x = -5 and x= 1 .. but it is supposed to be x=5 and x=1.

Solve:
x-2√(x-1) = 1


(x-2√(x-1))² = (1)²

x² +4(x-1)=1
x² + 4x - 4 =1
x²- +4x -5 = 0
(x+5)(x -1 )=0

x = -5, x = 1

agghh D:

I would probably do it this way....

x-2√(x-1) = 1

first get the square root stuff on one side. (i just think it's less complicated that way). so, subtract the x.

-2√(x-1) = 1 - x

now divide by -2.


√(x-1) = (1 - x) / -2

Now square both sides to get rid of the square root.

x-1= [(1-x)/ -2]^2

To square the right side you will do (1-x)(1-x) {see trick at bottom} which is 1 - 2x + x^2 all over -2^2 which is 4.
So you have
(x-1) = (1-2x + x^2) / 4

Then put the 4 back on the other side by multiplying it to both sides.

4x-4 = 1 - 2x + x^2

move the left terms over.

0 = 5 -6x + x^2

rearange to make it easier to factor.

0 = x^2 - 6x + 5
0 = (x-5)(x-1)


x = 5 x = 1

Hope this helps you !! (;


i tried to explain it in detail ... haha!




{the trick for squaring a binomial quickly is step 1- square first. 2- multiply the two numbers and then multiply what you get by 2. 3- square the last }