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Re: If x > 0 and 2x2 + 6x = 8, then the average (arithmetic mean
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22 Oct 2018, 09:06
1
Carcass wrote:
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
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
If x > 0 and 2x2 + 6x = 8, then the average (arithmetic mean
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08 Jul 2021, 08:50
1
Theory: Average = \(\frac{Sum Of All The Values}{Total Number Of Values}\)
We need to find the average (arithmetic mean) of x + 2, 2x – 1, and x + 4 Total Number of Values= 3 Sum of All the Values= x+2 + 2x-1 + x + 4 = x + 2x + x + 2 -1 +4 = 4x + 5
Average = \(\frac{Sum Of All The Values}{Total Number Of Values}\) = \(\frac{4x+5}{3}\)
Let's solve for x now
\(x > 0\) and \(2x^2 + 6x = 8\) => \(2x^2 + 6x - 8 = 0\) Divide both sides by 2 we get \(x^2 + 3x - 4 = 0\) => \(x^2 - x + 4x - 4 = 0\) => \(x(x - 1) + 4(x - 1) = 0\) => \((x - 1) (x+ 4) = 0\) => x = 1, -4 Since, x > 0 => x = 1
=> Average = \(\frac{4x+5}{3}\) = \(\frac{4*1+5}{3}\) = \(\frac{9}{3}\) = 3
So, Answer will be B Hope it helps!
To learn more about Statistics watch the following video