# Solve the equation (x+2)^1/2-(2-x)^1/2=2.Solve the equation (x+2)^1/2-(2-x)^1/2=2.

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First, we'll impose the constraint of existence of the square roots:

x+2 >=0

x> =-2

2-x>=0

x=<2

The interval of admissible values for x is [-2 ; 2].

Now, we'll solve the equation.

We'll multiply both sides by the conjugate of the expression sqrt(x+2) +sqrt(2-x).

sqrt(x+2) - sqrt(2-x) = 2 (1)

x + 2 - 2 + x = 2[sqrt(x+2) +sqrt(2-x)]

We'll combine and eliminate like terms:

2x = 2[sqrt(x+2) +sqrt(2-x)]

We'll divide by 2:

[sqrt(x+2) +sqrt(2-x)] = x (2)

We'll add (1) + (2):

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

We'll combine and eliminate like terms:

2sqrt(x+2) = x+2

We'll square raise both sides:

4(x+2) = (x+2)^2

We'll subtract 4(x+2) both sides:

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

We'll factorize by (x+2):

(x+2)(x+2-4) = 0

We'll put each factor as zero:

x + 2 = 0

x = -2

x - 2 = 0

x = 2

Since the values for the roots are the endpoints of the interval of admissible values, we'll accept both.