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

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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}.**

To solve for (f(x))^2 = 4. for f(x) = |x+4|-3.

Since{ f(x) }^2= 4, f(x) = sqrt4 = 2. Or f(x) = -sqrt4 = -2.

case (1)

If f(x) = 2, then |x+4| -3 = 2.

If x > =-4, then |x+4)-3 = x+4 -3= 2, x = 2+3-4 = 1. So **x = 1.**

If x < -4, then |x+4)-3 = -(x+4) -3 = 2, or -x= 2+3+4 = 9, or **x = -9**.

Case 2

If |x+4|-3 = -2.

When x> -4, then |X+4|-3 = x+4-3 = -2. Or x = -2+3-4. Or **x = -3.**

x< -4, |x+4| -3 = -(x+4)-3 = -2. Or -x-4-3 = -2. So -x = -2+3+4 = 5. Or** x = -5.**

x = 1 or x= -9 or x = 3 or x= -5.