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

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Bunuel
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If x > 0 and 2x^2 + 6x = 8, then the average (arithmetic mean) of x + [#permalink] New post   22 Oct 2018, 21:35
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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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B
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B
tam87
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Re: If x > 0 and 2x^2 + 6x = 8, then the average (arithmetic mean) of x + [#permalink] New post   22 Oct 2018, 22:15
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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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Re: If x > 0 and 2x^2 + 6x = 8, then the average (arithmetic mean) of x + [#permalink] New post   23 Oct 2018, 06:05
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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

Answer: B

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Re: If x > 0 and 2x^2 + 6x = 8, then the average (arithmetic mean) of x + [#permalink] New post   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] New post   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
B is the answer

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Re: If x > 0 and 2x^2 + 6x = 8, then the average (arithmetic mean) of x + [#permalink] New post   30 Apr 2020, 03:47
Expert Reply
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

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
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Re: If x > 0 and 2x^2 + 6x = 8, then the average (arithmetic mean) of x + [#permalink] New post   05 Aug 2022, 23:03
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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!

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Re: If x > 0 and 2x^2 + 6x = 8, then the average (arithmetic mean) of x + [#permalink]
05 Aug 2022, 23:03
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