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-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!
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
To solve for w in the equation √(2 - w) = w, you can follow these steps:
Step 1: Square both sides of the equation to eliminate the square root symbol:
(√(2 - w))^2 = w^2
2 - w = w^2
Step 2: Rearrange the equation to isolate the quadratic term on one side:
w^2 + w - 2 = 0
Step 3: Factor the quadratic equation. In this case, the factored form is: (w + 2)(w - 1) = 0
Setting each factor equal to zero, you get two separate equations:
w + 2 = 0 or w - 1 = 0
Step 4: Solve each equation for w:
For w + 2 = 0, you subtract 2 from both sides:
w = -2
For w - 1 = 0, you add 1 to both sides:
w = 1
Therefore, the solutions for w in the equation √(2 - w) = w are w = -2 and w = 1.