√(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!
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
2 answers
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
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
4 answers