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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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?
square both sides ...
$\displaystyle (\sqrt{x - 2})^2 = 1^2$
... proceed.
deleting
wait i see my mistake...
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.
$\displaystyle (\sqrt{x-2})^2 = x-2$ , not $\displaystyle (x-2)(x-2)$
I made that more complicated then it needed to be. thanks.