# What is the solution of equation [square root(4x+8)] - x=3?

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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.**