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(k-5)^2=0

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(k-5)^2=0 [#permalink] New post   30 Jun 2020, 08:56
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\((k-5)^2=0\)

Quantity A
Quantity B
\([k+3][k-3]\)
\([k+5][k-5]\)



A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
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Re: (k-5)^2=0 [#permalink] New post   30 Jun 2020, 08:56
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Re: (k-5)^2=0 [#permalink] New post   01 Jul 2020, 05:12
(k-5)(k-5) = 0
therefore: (k-5) or (k-5) must be 0
k-5= 0 when k = 5
The sole solution is then k = 5

Quantity A
(5+3)(5-3) = (8)(2) = 16

Quantity B
(5+5)(5-5) = (10)(0) = 0

Answer = A
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Re: (k-5)^2=0 [#permalink]
01 Jul 2020, 05:12
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