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Director
Joined: 16 May 2014
Posts: 592
Given Kudos: 0
GRE 1: Q165 V161
GRE Math Challenge #59-best approximation of x
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28 Feb 2015, 02:39
28 Feb 2015, 02:39
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Question Stats:
65% (00:50) correct
34% (01:07) wrong based on 103 sessions
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If \(x = (\sqrt{5} - \sqrt{7})^2\), the best approximation of x is :
A. 0
B. 1
C. 2
D. 3
E. 4
A. 0
B. 1
C. 2
D. 3
E. 4
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
Re: GRE Math Challenge #59
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22 Apr 2015, 06:07
22 Apr 2015, 06:07
2
soumya1989 wrote:
If \(x = (\sqrt{5} - \sqrt{7})^2\), the best approximation of x is :
A. 0
B. 1
C. 2
D. 3
E. 4
A. 0
B. 1
C. 2
D. 3
E. 4
Let's expand this:
(√5 - √7)² = (√5 - √7)(√5 - √7)
= √25 - √35 - √35 + √49
= 5 - 2√35 + 7
= 12 - 2√35
NOTE: since √36 = 6, we can see that the value of √35 is pretty close to 6
So, we get...
≈ 12 - 2(6)
≈ 0
Answer:
Show: ::
A
Cheers,
Brent
Re: GRE Math Challenge #59-best approximation of x
[#permalink]
10 Aug 2016, 17:59
10 Aug 2016, 17:59
I may have misunderstood the rules to adding and subtracting square roots, but don't the - √35 - √35 eliminate one another?
