What is the solution of equation [square root(4x+8)] - x=3?
The equation to be solved is : [square root (4x+8)] - x=3
[square root (4x+8)] - x = 3
=> [square root (4x+8)] = 3 + x
take the square of both the sides
=> 4x + 8 = 9 + 6x + x^2
=> x^2 + 2x + 1 = 0
=> (x + 1)^2 = 0
=> x = -1
The solution of the equation is x = -1.
We'll start by imposing the constraints of existence of square root:
4x + 8 >= 0
We'll subtract 8:
4x >= -8
x >= -2
The interval of admissible values for x is: [-2 , +infinite).
We'll shift x to the right, to isolate the square root to the left.
Now, we'll solve the equation raising to square both sides:
4x + 8 = (x+3)^2
We'll expand the square from the right side:
4x + 8 = x^2 + 6x + 9
We'll shift all terms to the right side and then we'll apply the symmetric property:
x^2 + 6x + 9 - 4x - 8 = 0
We'll combine like terms:
x^2 + 2x + 1 = 0
We'll recognize a perfect square:
(x+1)^2 = 0
x1 = x2 = -1
Since the value of x belongs to the range [-2 , +infinite), we'll accept it as solution of the equation: x = -1.