# What is the value of x in x+6 = 6square root(x-2)?

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### 2 Answers

We have to solve x+6 = 6* sqrt (x-2)

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

square both the sides

(x + 6)^2 = 36( x - 2)

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

=> x^2 - 24x + 108 = 0

=> x^2 - 18x - 6x + 108 = 0

=> x(x - 18) - 6(x - 18) = 0

=> (x - 6)(x - 18) = 0

**Therefore x = 6 and x = 18**

Before solving 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.

It would be better to let 6 to the left side. Why? Because raising to square both sides, we'll eliminate the square root. If we'll move 6 to the right side, we'll have to raise to square twice, to eliminate the square root.

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

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

We'll remove the bracktes:

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 allowed interval, we'll validate them as solutions: **

**x1 = 18 and x2 = 6.**