Crazy Radical Equation

[nycmathguy]

Solve for x.

sqrt{2 - sqrt{x + 2}} = x

Hint: square root within a square root.

 

[MathLover1]

sqrt(2 - sqrt(x + 2)) = x...........square both sides

2 - sqrt(x + 2) = x^2

2 - x^2=sqrt(x + 2) .........square both sides

(2 - x^2)^2=x + 2

x^4 - 4x^2 + 4=x + 2

x^4 - 4x^2 + 4-x - 2=0

x^4 - 4x^2 -x + 2=0

(x - 2) (x + 1) (x^2 + x - 1) = 0

solutions:
x=2
x=-1
x^2 + x - 1 = 0=>x = sqrt(5)/2 - 1/2 or x = -1/2 - sqrt(5)/2

verify solutions:
x=2
sqrt(2 - sqrt(x + 2)) =2 ->false
x=-1
sqrt(2 - sqrt(x + 2)) =-1 ->false

x=sqrt(5)/2 - 1/2

sqrt(2 - sqrt(sqrt(5)/2 - 1/2 + 2)) =sqrt(5)/2 - 1/2 =>true

x = -1/2 - sqrt(5)/2
sqrt(2 - sqrt(-1/2 - sqrt(5)/2 + 2)) =-1/2 - sqrt(5)/2 =>false


solution that works:

x = sqrt(5)/2 - 1/2

 

[nycmathguy]

sqrt(2 - sqrt(x + 2)) = x...........square both sides

2 - sqrt(x + 2) = x^2

2 - x^2=sqrt(x + 2) .........square both sides

(2 - x^2)^2=x + 2

x^4 - 4x^2 + 4=x + 2

x^4 - 4x^2 + 4-x - 2=0

x^4 - 4x^2 -x + 2=0

(x - 2) (x + 1) (x^2 + x - 1) = 0

solutions:
x=2
x=-1
x^2 + x - 1 = 0=>x = sqrt(5)/2 - 1/2 or x = -1/2 - sqrt(5)/2

verify solutions:
x=2
sqrt(2 - sqrt(x + 2)) =2 ->false
x=-1
sqrt(2 - sqrt(x + 2)) =-1 ->false

x=sqrt(5)/2 - 1/2

sqrt(2 - sqrt(sqrt(5)/2 - 1/2 + 2)) =sqrt(5)/2 - 1/2 =>true

x = -1/2 - sqrt(5)/2
sqrt(2 - sqrt(-1/2 - sqrt(5)/2 + 2)) =-1/2 - sqrt(5)/2 =>false


solution that works:

x = sqrt(5)/2 - 1/2

Magnificent math work. Bravo!