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 Post subject: GRE Functions (Select One)
PostPosted: Sun Oct 23, 2011 12:55 pm 
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Joined: Sun May 30, 2010 3:15 am
Posts: 424
If y = (x – 5)² + (x + 1)² – 6, then y is least when x =

A. -2
B. -1
C. 0
D. 2
E. 5

(D) Let us transform the formula:
y = (x – 5)² + (x +1)² – 6 =
x² – 10x + 25 + x² + 2x + 1 – 6 =
2x² – 8x + 20 = 2 × (x² – 4x + 10) =
2 × ((x² – 4x + 4) + 6) =
2 × ((x – 2)² + 6)

Any square is greater or equal 0. Therefore the formula possess the least value when (x – 2)² = 0.
x – 2 = 0
x = 2
The correct answer is choice (D).
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Can you please explain how the +10 was broken down into the +4) + 6? I was able to get to 2 × (x² - 4x+10) but not any further.


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 Post subject: Re: t.3, qt.10: algebra
PostPosted: Sun Oct 23, 2011 1:02 pm 
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Joined: Sun May 30, 2010 2:23 am
Posts: 498
Quote:
+4) + 6?
This is the method to get (ax + b)² + c type of a formula.

Here we have x² – 4x + 10. Knowing the formula of a square of a sum, (a + b)² = a² + 2ab + b², we look at x² – 4x and look at it as x² – 2 × (2 × x). The missing b² must be 2² = 4.

Therefore we represent x² – 4x + 10 as (x² – 4x + 4) + 6.


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