Posted by **Kathy** on Wednesday, September 3, 2014 at 7:20pm.

solve for w: the square root of 2-w=w

2-w=w^

w^+w-2

(w+2) (w-10)

w+2=0 w-1=0

w=-2 w=1

- algebra -
**Steve**, Wednesday, September 3, 2014 at 7:27pm
√(2-w) = w^2

2-w = w^4

w^4+2w-2 = 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)(w-10)

I'm sure you meant

(w+2)(w-1) = 0

In which case the roots are indeed -2 and 1

You seriously need to improve you exponential notation!

- algebra -
**Reiny**, Wednesday, September 3, 2014 at 7:28pm
Wow, that is some creative algebra

I assume your equation is

√(2-w) = 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

(√(2-w) )^2 = w^2

2 - w = w^2

w^2 + w - 2 = 0

(w+2)(w-1) = 0

w = -2 or w = 1

Since we squared, both solutions MUST be verified

if x = 1

LS = √(2-1) = 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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