# What is the value of x in 2x+12*[1 - (x-2)^(1/2)]=0?

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Before solving aÂ square root equation, we'll have to impose the constraint of existence of the square root.

The radicand has to be positive:

x - 2>=0

x>=2

So, all the solutions of the equation have to belong to the interval [2;+infinite).

Now, we'll solve the equation. We'll divide by 2:

x + 6 - 6sqrt(x-2) = 0

We'll move - 6sqrt(x-2) to the right side, so that raising to square both sides, we'll eliminate the square root.

(x+6)^2 = [6square root(x-2)]^2

x^2 + 12x + 36 = 36(x-2)

We'll remove the brackets:

x^2 + 12x + 36 - 36x + 72 = 0

We'll combine like terms:

x^2 - 24x + 108 = 0

x1 = [24+sqrt(144)]/2

x1 = (24+12)/2

x1 = 18

x2 = 6

**Since both values are in the interval of possible values, we'll validate them as solutions: x1 = 18 and x2 = 6.**