Solve the equation [f(x)]^2 = 4 for f(x)=|x+4|-3?

We'll start by explaining the modulus
|x+4|.

Case 1: |x+4| = x + 4 for
x+4>=0

We'll solve the inequality
x+4>=0:

x+4>=0

x
>= -4

The interval of admissible values for x is:
[-4,+infinite).

Case 2: |x+4| = -x - 4 for
x+4<0

x<-4

The
interval of admissible values for x is: (- infinite,
-4).

Now, we'll solve the equation in both
cases:

1) [f(x)]^2 = 4 for 
x>=-4

f(x) = x + 4 -
3

f(x) = x + 1

(x+1)^2 =
4

We'll expand the square:

x^2
+ 2x + 1 - 4 = 0

x^2 + 2x - 3 =
0

We'll apply the quadratic
formula:

x1 = [-2+sqrt(4 +
12)]/2

x1 = (-2+4)/2

x1 =
1

x2 = -3

Since both solutions
belong to the interval [-4,+infinite), they are
accepted.

2)  [f(x)]^2 = 4 for 
x<-4

f(x) = -x - 4 -
3

f(x) = -x - 7

(-x - 7)^2 =
4

We'll expand the square:

x^2
+ 14x + 49 - 4 = 0

x^2 + 14x + 45 =
0

We'll apply the quadratic
formula:

x1 =
[-14+sqrt(196-180)]/2

x1 =
(-14+4)/2

x1 = -5

x2 =
-9

Since both solutions belong to the interval (- infinite,
-4), they are accepted.

The equation [f(x)]^2
= 4 has the solutions: {-9;-5;-3;1}.