STRATEGY: As with all GRE Multiple Choice questions, we should immediately ask ourselves, Can I use the answer choices to my advantage?
In this case, we can test each answer choice by plugging it into the function to see which value satisfies the equation f(2k) = 36.
So, that's one option.
Now we should give ourselves 15-20 seconds to identify a faster approach.
In this case, we can also solve the equation.
I'm pretty sure I can quickly solve the equation so I'm going to go with that approach.


If \(f(x) = x^2 + 4\), then \(f(2k) = (2k)^2 + 4= 4k^2 + 4\)

Since we are told \(f(2k) = 36\), we can write: \(4k^2 + 4 = 36\)
Divide both sides of the equation by \(4\) to get: \(k^2 + 1 = 9\)
Subtract \(1\) from both sides to get: \(k^2 = 8\)

So, EITHER \(k = \sqrt{8} \) OR \(k = -\sqrt{8} \)
Check the answer choices...... Not there. It looks like we need to simplify \(\sqrt{8}\)

\(\sqrt{8} = \sqrt{4 \times 2} = \sqrt{4} \times \sqrt{2}= 4 \sqrt{2}\).
So, EITHER \(k = 2\sqrt{2} \) OR \(k = -2\sqrt{2} \)

Answer: D

Given that f(x) = \(x^2\) + 4 and f(2k) = 36 and we need to find one possible value of k

To find f(2k) we need to compare what is inside the bracket () * in f(2k) and f(x)

=> We need to substitute x with 2k in f(x) = \(x^2\) + 4 to get the value of f(2k)

=> f(2k) = \((2k)^2\) + 4 = \(4k^2\) + 4 = 36 (given)
=> \(4k^2\) = 36-4 = 32
=> \(k^2\) = \(\frac{32}{4}\) = 8
=> k = \(\sqrt{8}\) = \(2\sqrt{2}\)

So, Answer will be D
Hope it helps!

Watch the following video to learn the Basics of Functions and Custom Characters

If \(f(x)=x^2+4\), and \(f(2k)=36\). then which of the following is one possible value of K?

A. \(\sqrt{2}\)

B. 2

C. 4

D. \(2 \sqrt{2}\)

E. \(\sqrt{14}\)