If f(x) = x² - 2x and g(x) = 6x + 4, for which value of x does (f + g)(x) = 0?
-4, -2, 2, 4

Solution:

Given are two functions f(x) and g(x).

To find: (f + g)(x) = 0

We know that f(x) + g(x) = (f + g)(x)

Step 1: Solve f(x) + g(x)

∴ (f + g) (x) = (x² - 2x) + (6x + 4)

⇒ x² + 4x + 4 = 0

Step 2: Find the solutions of the equation (f + g) (x) = 0.

x² + 4x + 4 = 0

In the above equation, a = 1, b = 4 and c = 4

Step 3: Split the middle term.

⇒ x² + 2x + 2x + 4 = 0

⇒x(x + 2) + 2 (x + 2) = 0

⇒ (x + 2)(x + 2) = 0

Step 4: Equate both the factors to 0.

x = -2

If f(x) = x² - 2x and g(x) = 6x + 4, for which value of x does (f + g)(x) = 0?

Summary:

The value of x = - 2 with the multiplicity of 2 as it occurs twice as the factors, at (f + g)(x) = 0