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If q = 43 + (p - 7)^2, then q is lowest when p =

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Bunuel
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If q = 43 + (p - 7)^2, then q is lowest when p = [#permalink] New post   13 Nov 2014, 09:46
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A
B
C
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E

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5% (low)

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84% (00:41) correct 16% (01:01) wrong based on 170 sessions
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Tough and Tricky questions: Min/Max Problems.



If \(q = 43 + (p - 7)^2\), then \(q\) is lowest when \(p =\)

A. 0
B. 7
C. 10
D. 14
E. 43

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B
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kinghyts
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Re: If q = 43 + (p - 7)^2, then q is lowest when p = [#permalink] New post   13 Nov 2014, 20:35
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q = 43 + (p - 7)^2 is least when (p-7)^2 is least i.e p-7 = 0
Therefore the answer is b
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sunaimshadmani
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Re: If q = 43 + (p - 7)^2, then q is lowest when p = [#permalink] New post   13 Nov 2014, 09:52
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q = 43 + (p - 7)^2, in this expression (p-7) is a square whose minimum value is zero, so if p=7, q is minimal. Answer is B.
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NATABERI
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Re: If q = 43 + (p - 7)^2, then q is lowest when p = [#permalink] New post   13 Nov 2014, 10:05
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the smallest possible answer of (p - 7)^2 is 0 , when p=7
(7-7)^2=0^2=0

answer: B
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Re: If q = 43 + (p - 7)^2, then q is lowest when p = [#permalink] New post   13 Nov 2014, 12:43
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easy peasy, just plug away ^_^

A. plug in 0 to get 43 + (-7)^2 = 43 + 49 = 92
B. 43 + (0) ^ 2 = 43 + 0 = 43
C. 43 + (3) ^2 = 43 + 9 = 52
D. 43 + (7) ^ 2 = 43 + 49 = 92
E. 43 + (36) ^2 = no need to calculate haha this is way too big

Answer choice B is the smallest!
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Re: If q = 43 + (p - 7)^2, then q is lowest when p = [#permalink] New post   13 Nov 2014, 20:14
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Answer = B = 7

43 is a constant. There is a +ve sign after it.

To reduce 43, we require to get the next expression -ve.

However, square of any number is +ve. Maximum the expression can be equated is to zero

\((p-7)^2 = 0\)

p = 7
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Re: If q = 43 + (p - 7)^2, then q is lowest when p = [#permalink] New post   14 Nov 2014, 05:49
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Let's expand the equation, then it will be

q=43+\(p^{2}\)-14p+49, Compare this equation with general equation a\(x^{2}\)+bx+c

a=1 and b=-14

q is lowest when p =-b/2a = -(-14)/2 =7

Hence B.
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Re: If q = 43 + (p - 7)^2, then q is lowest when p = [#permalink] New post   14 Nov 2014, 09:28
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Bunuel wrote:

Tough and Tricky questions: Min/Max Problems.



If \(q = 43 + (p - 7)^2\), then \(q\) is lowest when \(p =\)

A. 0
B. 7
C. 10
D. 14
E. 43

Kudos for a correct solution.


If \(q = 43 + (p - 7)^2\), then \(q\) is lowest when \(p =\)

A. 0
B. 7
C. 10
D. 14
E. 43

The question asks us for the value of \(p\) that yields the lowest value of \(q\).

Since the quantity \((p - 7)\) is squared, this value will always be non-negative. Therefore, \(q\) is lowest when \((p - 7)^2 = 0\).

Only one number when squared is equal to zero: zero itself. Thus \((p - 7) = 0\). Adding 7 to both sides gives \(p = 7\).

Therefore, \(q\) is lowest when \(p = 7\).

Answer: B.
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Re: If q = 43 + (p - 7)^2, then q is lowest when p = [#permalink] New post   14 Jul 2022, 07:37
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Re: If q = 43 + (p - 7)^2, then q is lowest when p = [#permalink]
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