BTGModeratorVI wrote: ↑

Sun Apr 05, 2020 9:03 am

If x > 0 and 2x^2 + 6x = 8, then the average (arithmetic mean) of x + 2, 2x – 1, and x + 4 is equal to which of the following?

A. –2
B. 3
C. 3.5
D. 5
E. 7

Answer: B
Source: Kaplan

Given: 2x² + 6x = 8
Divide both sides by 2 to get: x² + 3x = 4
Subtract 4 from both sides to get: x² + 3x - 4 = 0
Factor: (x + 4)(x - 1) = 0
Solved to get: EITHER x = -4 OR x = 1

Since we're told that x > 0, we can be certain that x = 1

QUESTION: What is the average (arithmetic mean) of x + 2, 2x – 1, and x + 4?
If x = 1, then:
x + 2 = 1 + 2 = 3
2x - 1 = 2(1) - 1 = 1
x + 4 = 1 + 4 = 5

Average of 3, 1 and 5 = (3 + 1 + 5)/3
= 9/3
= 3

Answer: B

Cheers,
Brent

BTGModeratorVI wrote: ↑

Sun Apr 05, 2020 9:03 am

If x > 0 and 2x^2 + 6x = 8, then the average (arithmetic mean) of x + 2, 2x – 1, and x + 4 is equal to which of the following?

A. –2
B. 3
C. 3.5
D. 5
E. 7

Answer: B
Source: Kaplan

Let’s solve the given equation:

2x^2 + 6x - 8 = 0

x^2 + 3x - 4 = 0

(x - 1)(x + 4) = 0

x = 1 or x = -4

Since we are given that x > 0, then x = 1. So the average of x + 2, 2x – 1, and x + 4 is equal to

(3 + 1 + 5)/3 = 9/3 = 3

Answer: B