# Solve for x. f(g(x))=0 f(x)=x^2-4 and g(x)=(x+3)

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We are given that f(x) = x^2 - 4 and g(x) = x + 3. We have to solve for x given that f(g(x)) = 0.

f(g(x)) = 0

=> f( x +3) = 0

=> (x +3)^2 - 4 = 0

=> (x + 3 - 2)(x + 3 + 2) = 0

(x + 3 - 2) = 0

=> x = -1

(x + 3 + 2) = 0

=> x = -5

**The solutions for x are x = -1 and x = -5**

We'll substitute g(x) by it's expression:

f(g(x)) = f((x+3))

f((x+3)) = (x+3)^2 - 4

We'll expand the square and we'll get:

f((x+3)) = x^2 + 6x + 9 - 4

f((x+3)) = x^2 + 6x + 5

We'll solve the equation:

f((x+3)) = 0 <=> x^2 + 6x + 5 = 0

We'll apply the quadratic formula:

x1 = [-6+sqrt(36 - 20)]/2

x1 = (-6 + 4)/2

x1 = -1

x2 = -5

**The solutions of the equation f(g(x))=0 are: {-1 ; -5}.**