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

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Joined: 02 Sep 2009
Posts: 58383
If x > 0 and 2x^2 + 6x = 8, then the average (arithmetic mean) of x +  [#permalink]

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22 Oct 2018, 21:35
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25% (medium)

Question Stats:

77% (01:37) correct 23% (01:54) wrong based on 96 sessions

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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

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Joined: 28 Mar 2018
Posts: 5
Re: If x > 0 and 2x^2 + 6x = 8, then the average (arithmetic mean) of x +  [#permalink]

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22 Oct 2018, 22:15
1
2x^2 + 6x - 8 = 0
or, x^2 + 3x - 4 = 0
or, (x + 4) (x - 1) = 0
As x > 0, x = 1

mean of the terms = 1/3 [x + 2 + 2x - 1 + x + 4] = 1/3 [4x + 5] = 1/3 [9] = 3 (Choice B)
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Joined: 12 Sep 2015
Posts: 4006
Re: If x > 0 and 2x^2 + 6x = 8, then the average (arithmetic mean) of x +  [#permalink]

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23 Oct 2018, 06:05
Top Contributor
Bunuel 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 2x^2 + 6x = 8, then the average (arithmetic mean) of x +  [#permalink]

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23 Oct 2018, 06:46
Bunuel 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

$$2x^2 + 6x = 8$$

Solve the Quadratic Equation, we have $$x = 1$$ , $$-4$$ ; further since it is given $$x > 0$$ , we are left with x = 1

Average of $$x + 2$$, $$2x – 1$$, and $$x + 4$$ is $$\frac{( x + 2 ) + ( 2x – 1 ) + ( x + 4 )}{3} = \frac{4x + 5}{3} = \frac{9}{3} = 3$$, Answer must be (B) 3
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Re: If x > 0 and 2x^2 + 6x = 8, then the average (arithmetic mean) of x +  [#permalink]

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23 Oct 2018, 07:14
2x^2 + 6x = 8
x^2 + 3x = 4
(x-1) (x+4) = 0
Its given in the ques x>0
Therefore, x = 1

Now (x+2)+(2x-1)+(x+4)/3
(4x+5)/3
Putting x = 1
4(1)+5/3
3

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Re: If x > 0 and 2x^2 + 6x = 8, then the average (arithmetic mean) of x +   [#permalink] 23 Oct 2018, 07:14
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