For which value of x does the equation f(g(x)) = 0 hold true, given the functions f(x) = x^2 - 2x and g(x) = 6x + 4?

To find the value of x for which f(g(x)) = 0, we first need to understand the given functions:

Next, we can substitute g(x) into f(x):

f(g(x)) = f(6x + 4)

Now, we replace x in the function f(x) with g(x):

f(g(x)) = (6x + 4)² - 2(6x + 4)

Expanding this expression:

f(g(x)) = (6x + 4)(6x + 4) - 2(6x + 4)

This further simplifies to:

f(g(x)) = 36x² + 48x + 16 - 12x - 8

Combining like terms, we get:

f(g(x)) = 36x² + 36x + 8

To find the value of x that makes this equal to zero, we set:

36x² + 36x + 8 = 0

Now we can use the quadratic formula, which states that for any equation of the form ax² + bx + c = 0, the solutions for x are given by:

x = (-b ± √(b² - 4ac)) / (2a)

In our case, a = 36, b = 36, and c = 8. Plugging these values into the quadratic formula:

x = (-36 ± √(36² - 4 * 36 * 8)) / (2 * 36)

Calculating the discriminant:

36² - 4 * 36 * 8 = 1296 - 1152 = 144

Now compute the solutions:

x = (-36 ± √144) / 72

x = (-36 ± 12) / 72

Calculating both potential values:

x = (-36 + 12) / 72 = -24 / 72 = -1/3

x = (-36 - 12) / 72 = -48 / 72 = -2/3

Therefore, the solutions are:

These are the values of x for which f(g(x)) = 0.

For which value of x does the equation f(g(x)) = 0 hold true, given the functions f(x) = x^2 – 2x and g(x) = 6x + 4?