# Find n for (4n+8)^1/2=n+3.

### 2 Answers | Add Yours

We have to find n : (4n+8)^1/2 = n + 3

(4n+8)^1/2 = n + 3

square both the sides

=> 4n + 8 = (n + 3)^2

=> 4n + 8 = n^2 + 9 + 6n

=> n^2 + 9 + 6n - 4n - 8 = 0

=> n^2 + 2n + 1 =0

=> (n +1)^2 = 0

=> n = -1

**Therefore n only has the roots n = -1.**

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

4n + 8 >= 0

We'll subtract 8:

4n >= -8

n >= -2

The range of admissible values for n is: [-2 , +infinite).

Now, we'll solve the equation raising to square both sides:

4n + 8 = (n+3)^2

We'll expand the square from the right side:

4n + 8 = n^2 + 6n + 9

We'll move all terms to the right side and we'll use the symmetric property:

n^2 + 6n + 9 - 4n - 8 = 0

We'll combine like terms:

n^2 + 2n + 1 = 0

(n+1)^2 = 0

n = -1

**Since n is not an extraneous solution, we'll state that the solution of the equation is x = -1.**