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If m^2 = 17, then what is the value of (m+1)(m−1)?

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If m^2 = 17, then what is the value of (m+1)(m−1)?  [#permalink]

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25 Nov 2018, 23:49
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15% (low)

Question Stats:

91% (00:24) correct 9% (01:10) wrong based on 40 sessions

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If m^2 = 17, then what is the value of (m+1)(m−1)?

A. √17 − 1
B. √17 + 1
C. 16
D. 18
E. 288

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Location: India
Concentration: Finance, Marketing
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If m^2 = 17, then what is the value of (m+1)(m−1)?  [#permalink]

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25 Nov 2018, 23:55
1
$$m^2$$ = 17
m = $$\sqrt{17}$$

(m+1)(m-1) = $$m^2 - 1^2$$ = 17-1 = 16

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Re: If m^2 = 17, then what is the value of (m+1)(m−1)?  [#permalink]

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25 Nov 2018, 23:58
m2 = 17
m2 - 1 = 17-1
(m-1)(m+1)= 16
C is correct
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Re: If m^2 = 17, then what is the value of (m+1)(m−1)?  [#permalink]

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26 Nov 2018, 05:32
option C.
m^2=17
m=sqrt17
(m+1)(m-1)=m^2-1^2=17-1=16
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Re: If m^2 = 17, then what is the value of (m+1)(m−1)?  [#permalink]

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26 Nov 2018, 08:57
C.
m2 = 17
(m-1)(m+1)=m2-1
= 17-1
=16
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Re: If m^2 = 17, then what is the value of (m+1)(m−1)?  [#permalink]

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26 Nov 2018, 21:28
Bunuel wrote:
If m^2 = 17, then what is the value of (m+1)(m−1)?

A. √17 − 1
B. √17 + 1
C. 16
D. 18
E. 288

(m+1)(m-1)= (m^2-1)

(17-1)= 16 option C
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Re: If m^2 = 17, then what is the value of (m+1)(m−1)?   [#permalink] 26 Nov 2018, 21:28
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