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Re: Which of the following is equal to the average (arithmetic mean) of (x [#permalink]
Answer: C

The arithmetic mean is the sum of the elements divided by the number of elements.
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Re: Which of the following is equal to the average (arithmetic mean) of (x [#permalink]
The sum of (x+2)^2 and (x-2)^2 = 2x^2+8
Hence Average = x^2+4


Hence C
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Re: Which of the following is equal to the average (arithmetic mean) of (x [#permalink]
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jsphcal wrote:
Which of the following is equal to the average (arithmetic mean) of (x+2)^2 and (x-2)^2?

A. x^2
B. x^2+2
C. x^2 +4
D. x^2+2x
E. x^2+4x


AM of (x+2)^2 +(x-2)^2

= \(\frac{( x^2 + 4x + 4 ) + ( x^2 - 4x + 4 )}{2}\)

= \(\frac{2x^2 + 8}{2}\)

= \(x^2 + 4\)

Hence, answer will be (C) \(x^2 + 4\)
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Re: Which of the following is equal to the average (arithmetic mean) of (x [#permalink]
Abhishek009 wrote:
jsphcal wrote:
Which of the following is equal to the average (arithmetic mean) of (x+2)^2 and (x-2)^2?

A. x^2
B. x^2+2
C. x^2 +4
D. x^2+2x
E. x^2+4x


AM of (x+2)^2 +(x-2)^2

= \(\frac{( x^2 + 4x + 4 ) + ( x^2 - 4x + 4 )}{2}\)

= \(\frac{2x^2 + 8}{2}\)

= \(x^2 + 4\)

Hence, answer will be (C) \(x^2 + 4\)

please I found answer B. because I factorized the expression (x+2)^2 = (x+2)(x-2) instead of (x+2)^2=(x-2)(x-2)
I used the algeabric formula of (a-b)^2=(a-b)(a+b) so my question is, how do I recognize when an expression of (a-b)^2 is equals to (a-b)(a+b) ?
Thank you in advance
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Re: Which of the following is equal to the average (arithmetic mean) of (x [#permalink]
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The equations should be valid for all value of X .. so let x=0 in the equations..
We get (0+2)^2=4 and (0-2)^2=4
Average of these two numbers =4

Only Ans (C) gives us this value

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Re: Which of the following is equal to the average (arithmetic mean) of (x [#permalink]
\(x^{2} + 4 + 2X + X^{2} + 4 -2X\)

\(\frac{2X^{2} + 8}{2}\)

\(\frac{2}{2} (X^{2} + 4)\)

C
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Re: Which of the following is equal to the average (arithmetic mean) of (x [#permalink]
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Re: Which of the following is equal to the average (arithmetic mean) of (x [#permalink]
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