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# If x > 0 and 2x2 + 6x = 8, then the average (arithmetic mean

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If x > 0 and 2x2 + 6x = 8, then the average (arithmetic mean [#permalink]  22 Oct 2018, 03:58
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Question Stats:

82% (01:05) correct 17% (01:06) wrong based on 28 sessions
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
[Reveal] Spoiler: OA

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Re: If x > 0 and 2x2 + 6x = 8, then the average (arithmetic mean [#permalink]  22 Oct 2018, 09:06
Expert's post
Carcass Array[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

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

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Re: If x > 0 and 2x2 + 6x = 8, then the average (arithmetic mean   [#permalink] 22 Oct 2018, 09:06
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