# solve square root equation x=6[squareroot(x-2)-1]?

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We'll impose the constraint of existence of the square root:

x-2>=0

We'll add 2 both sides:

x>=2

The interval of admissible values is [2 ; +infinite).

We'll re-write the given equation. We'll isolate 6sqrt(x-2) to the right side. For this reason, we'll add 6 both sides:

6+x = 6sqrt(x-2)

We'll square raise both sides. For squaring the binomial from the left side, we'll use the formula;

(a+b)^2 = a^2 + 2ab + b^2

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

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

We'll subtract 36x-72:

x^2 - 24x + 108 = 0

We'll apply the quadratic formula:

x1=[24+ sqrt (24^2-4*108)]/2

x1 = (24+12)/2

x1 = 36/2

x1=18

x2=(24-12)/2

x2=6

**Since both values belong to the interval [2 ; +infinite), we'll validate them as solutions of the given equation: x1 = 18 and x2 = 6.**