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If y = x^2 − 32x + 256, then what is the least possible value of y ?

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If y = x^2 − 32x + 256, then what is the least possible value of y ?  [#permalink]

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New post 14 Mar 2019, 00:18
00:00
A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

56% (01:26) correct 44% (01:32) wrong based on 32 sessions

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If y = x^2 − 32x + 256, then what is the least possible value of y ?  [#permalink]

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New post 14 Mar 2019, 00:23
\(y = x^2 − 32x + 256 = (x - 16)^2\)

At x=16, the value of y is 0

Therefore, the least value of y from the options available is 0(Option E)
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Re: If y = x^2 − 32x + 256, then what is the least possible value of y ?  [#permalink]

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New post 14 Mar 2019, 03:57
Bunuel wrote:
If y = x^2 − 32x + 256, then what is the least possible value of y ?

A. 256
B. 32
C. 16
D. 8
E. 0

y = x^2 − 32x + 256

(x-16)*(x-16)=0
x=16
so at x=16
we get y=0
IMO E
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Re: If y = x^2 − 32x + 256, then what is the least possible value of y ?  [#permalink]

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New post 18 Mar 2019, 18:49
Bunuel wrote:
If y = x^2 − 32x + 256, then what is the least possible value of y ?

A. 256
B. 32
C. 16
D. 8
E. 0


Since x^2 − 32x + 256 = (x - 16)(x - 16) = (x - 16)^2, we see that the least possible value of y is 0 (occuring at x = 16).

Answer: E
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Re: If y = x^2 − 32x + 256, then what is the least possible value of y ?   [#permalink] 18 Mar 2019, 18:49
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