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# Help needed hard algebra question

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The question

Mar 6, 2023

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We need to find the number of values of x that satisfy the equation f(f(x)) = 5. Let's start by solving this equation.

Since f(x) is defined piecewise, we need to consider two cases:

Case 1: x < -4 In this case, f(x) = x + 3. Therefore, f(f(x)) = f(x + 3) = (x + 3) - 3 = x - 6. So we need to solve the equation f(f(x)) = x - 6 = 5, which gives x = 11.

Case 2: x >= -4 In this case, f(x) = x^2 - 4. Therefore, f(f(x)) = f(x^2 - 4) = (x^2 - 4)^2 - 4. So we need to solve the equation f(f(x)) = (x^2 - 4)^2 - 4 = 5. Expanding the left side, we get (x^4 - 8x^2 + 12) = 0. This is a quartic equation that can be factored as (x^2 - 2)(x^2 - 6) = 0. Therefore, the solutions are x = sqrt(2), -sqrt(2), sqrt(6), and -sqrt(6).

We need to check which of these solutions also satisfy the condition x >= -4 since this is the second case of the piecewise definition of f(x).

We have:

sqrt(6) < -2 < -4, so -sqrt(6) is not a valid solution. sqrt(6) > 2 > -4, so sqrt(6) is a valid solution.

Therefore, there are two values of x that satisfy the equation f(f(x)) = 5: 11 and sqrt(6).

Mar 6, 2023