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# help

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Solve $$\dfrac{2}{x + \sqrt{2 - x^2}} + \dfrac{2}{x - \sqrt{2 - x^2}} = x$$

Dec 15, 2019

#1
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Solve for x:
2/(x - sqrt(2 - x^2)) + 2/(x + sqrt(2 - x^2)) = x

Bring 2/(x - sqrt(2 - x^2)) + 2/(x + sqrt(2 - x^2)) together using the common denominator (sqrt(2 - x^2) - x) (x + sqrt(2 - x^2)):
-(4 x)/((sqrt(2 - x^2) - x) (x + sqrt(2 - x^2))) = x

Multiply both sides by (sqrt(2 - x^2) - x) (x + sqrt(2 - x^2)):
-4 x = x (sqrt(2 - x^2) - x) (x + sqrt(2 - x^2))

x (sqrt(2 - x^2) - x) (x + sqrt(2 - x^2)) = 2 x - 2 x^3:
-4 x = 2 x - 2 x^3

Subtract 2 x - 2 x^3 from both sides:
2 x^3 - 6 x = 0

Factor x and constant terms from the left hand side:
2 x (x^2 - 3) = 0

Divide both sides by 2:
x (x^2 - 3) = 0

Split into two equations:
x = 0 or x^2 - 3 = 0

x = 0 or x^2 = 3

Take the square root of both sides:

x = 0    or    x = sqrt(3)    or    x = -sqrt(3)

Dec 15, 2019
#2
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Get a commond denominator on the  left and we have that

2 [ x - √(2 - x^2) ]  + 2 [ x +  √(2 - x^2) ]

______________________________   =  x

x^2   - (2 - x^2)

4x

________   =    x

2x^2 - 2

2x

_______  = x

x^2  - 1

2x = x^3 - x

x^3 - 3x   = 0

x ( x^2 - 3)  =  0

(x- 0)(x - √3) ( x + √3)  =  0

Setting each factor to 0  and solving for x gives the solutions

x =0, x = √3 and x = -√3   Dec 15, 2019