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Difficulty: HardFunction Evaluation, Composition, and Properties

For all real numbers x≥−5x \geq -5, the functions ff and gg are defined by f(x)=x2−4xf(x) = x^2 - 4x and g(x)=x+5g(x) = \sqrt{x + 5}. If f(g(k))=12f(g(k)) = 12, what is the real value of kk?

Answer: 31

Answer

The correct answer is 31.
Substituting g(k)=k+5g(k) = \sqrt{k+5} into f(x)=x2−4xf(x) = x^2 - 4x gives the equation (k+5)2−4k+5=12(\sqrt{k+5})^2 - 4\sqrt{k+5} = 12. Simplifying and isolating the radical yields k−7=4k+5k - 7 = 4\sqrt{k+5}. Squaring both sides results in (k−7)2=16(k+5)(k-7)^2 = 16(k+5), which simplifies to k2−30k−31=0k^2 - 30k - 31 = 0. Factoring this equation gives (k−31)(k+1)=0(k-31)(k+1) = 0, yielding potential solutions of k=31k = 31 and k=−1k = -1. Checking these solutions reveals that k=−1k = -1 is extraneous because f(g(−1))=−4≠12f(g(-1)) = -4 \neq 12. Therefore, the only valid real solution is k=31k = 31.

Step-by-Step Solution

1
Express the composition f(g(k))f(g(k)) using the given functions.
f(g(k))=(g(k))2−4(g(k))f(g(k)) = (g(k))^2 - 4(g(k))
To evaluate a composite function, substitute the inner function g(k)g(k) as the input into the outer function ff.
2
Substitute g(k)=k+5g(k) = \sqrt{k+5} and set the composite expression equal to 12.
(k+5)2−4k+5=12(\sqrt{k+5})^2 - 4\sqrt{k+5} = 12
This sets up the equation to solve for the unknown variable kk.
3
Isolate the radical term on one side of the equation.
k−7=4k+5k - 7 = 4\sqrt{k+5}
Simplifying (k+5)2(\sqrt{k+5})^2 to k+5k+5 and moving terms helps isolate the radical before squaring.
4
Square both sides of the equation to eliminate the square root.
(k−7)2=16(k+5)(k-7)^2 = 16(k+5)
Squaring is the inverse operation of the square root, which removes the radical.
5
Expand both sides and rewrite the equation in standard quadratic form.
k2−30k−31=0k^2 - 30k - 31 = 0
Expanding (k−7)2(k-7)^2 to k2−14k+49k^2 - 14k + 49 and 16(k+5)16(k+5) to 16k+8016k + 80, then moving all terms to one side, allows us to solve the resulting quadratic equation.
6
Factor the quadratic equation.
(k−31)(k+1)=0(k-31)(k+1) = 0
Factoring allows us to find the potential roots easily.
7
Solve for the potential values of kk.
k=31k = 31 or k=−1k = -1
Setting each factor to zero gives the candidate solutions.
8
Verify both potential values in the original equation to check for extraneous solutions.
k=31k = 31 is valid; k=−1k = -1 is extraneous.
Squaring both sides can introduce extraneous solutions. Evaluating f(g(−1))f(g(-1)) yields −4-4, not 1212, while evaluating f(g(31))f(g(31)) yields 1212.

Key Concept

Function composition involves substituting one function into another, and solving equations with radicals requires checking for extraneous solutions.
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Topics

Function Evaluation, Composition, and Properties

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