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
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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}.
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