Needing some review with solving this

SQRT(x-2)-1 = 0

I carried the 1 over making

SQRT(x-2)= 1

at this point do I want to carry the 2 over as well?

or do I square the 1 right away?

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- Oct 16th 2012, 02:57 PMwolfwoodSQRT(x-2)-1 = 0
Needing some review with solving this

SQRT(x-2)-1 = 0

I carried the 1 over making

SQRT(x-2)= 1

at this point do I want to carry the 2 over as well?

or do I square the 1 right away? - Oct 16th 2012, 03:03 PMskeeterRe: SQRT(x-2)-1 = 0
square both sides ...

$\displaystyle (\sqrt{x - 2})^2 = 1^2$

... proceed. - Oct 16th 2012, 03:07 PMwolfwoodRe: SQRT(x-2)-1 = 0
deleting

- Oct 16th 2012, 03:07 PMwolfwoodRe: SQRT(x-2)-1 = 0
wait i see my mistake...

- Oct 16th 2012, 03:14 PMwolfwoodRe: SQRT(x-2)-1 = 0
if the next step is (x-2)(x-2) = 1

I'm going to be left with

X^2 -4x + 3 = 0

This ties back to my other post when im solving f(x) = 0 for 2 < x < 11

This doesn't seem like it could be solved.

inless

i simplify it to (1 + -1x)(3 + -1x) = 0

X = - 1 or X = -3

Did i do that right?

Both wouldnt fit into the interval. - Oct 16th 2012, 03:24 PMskeeterRe: SQRT(x-2)-1 = 0
$\displaystyle (\sqrt{x-2})^2 = x-2$ , not $\displaystyle (x-2)(x-2)$

- Oct 16th 2012, 03:25 PMwolfwoodRe: SQRT(x-2)-1 = 0
I made that more complicated then it needed to be. thanks.