sqrt 3x+ 18 =x
3(6) + 18 = x
18+ 18 = 36
36 = x
sqrt 36
=6
So the solution is 6.
Panama,
You cannot just “throw out” the “sqrt”, as you have done in your first step. No, the solution is not 6.
Also, you have not told us if the x is under the square root (also called the radical symbol) or not. Use parentheses to show this.
Note, also, that you can show the sqrt as an exponent equal to 1/2 (e.g., sqrt 5 = 5^.5).
I will assume that the x is under the sqrt with the 3. In that case, it can be written as either
Sqrt(3x) or (3x)^.5
So your problem goes like this:
(3x)^.5 + 18 = x Now subtract 18 from both sides.
(3x)^.5 = x – 18 Next square both sides of the equation.
((3x)^.5)^2 = (x – 18)(x – 18)
3x = x^2 – 36x + 324 Now subtract 3x from each side in order to make the left hand side equal zero.
0 = x^2 – 39x + 324
Now solve the quadratic.