algebra
posted by Kathy .
solve for w: the square root of 2w=w
2w=w^
w^+w2
(w+2) (w10)
w+2=0 w1=0
w=2 w=1

√(2w) = w^2
2w = w^4
w^4+2w2 = 0
The solutions to that are not trivial.
So, assuming you meant
√2  w = w
√2 = 2w
2 = 4w^2
w = ±1/√2
Better try again. Ignoring the first two lines, once you get to
(w+2)(w10)
I'm sure you meant
(w+2)(w1) = 0
In which case the roots are indeed 2 and 1
You seriously need to improve you exponential notation! 
Wow, that is some creative algebra
I assume your equation is
√(2w) = w , or is it √2  w = w
(notice how my brackets totally change the meaning of the problem)
I will assume the first
square both sides
(√(2w) )^2 = w^2
2  w = w^2
w^2 + w  2 = 0
(w+2)(w1) = 0
w = 2 or w = 1
Since we squared, both solutions MUST be verified
if x = 1
LS = √(21) = 1
RS = 1^2 = 1 = LS
so x = 1
if x = 2
LS = √(2 + 2) = √4 = 2
RS = 2
LS ≠ RS
so the only solution is x = 1
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