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16 Sep 2014, 00:13
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A
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Difficulty:
25%
(medium)
Question Stats:
70% (01:00) correct
30%
(01:17)
wrong
based on 239
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If \(x=(\sqrt{5}-\sqrt{7})^2\), then 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. \(4\)
16 Sep 2014, 00:14
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Official Solution:
If \(x=(\sqrt{5}-\sqrt{7})^2\), then the best approximation of \(x\) is:
A. \(0\)
B. \(1\)
C. \(2\)
D. \(3\)
E. \(4\)
\(x=(\sqrt{5}-\sqrt{7})^2=5-2\sqrt{35}+7=12-2\sqrt{35}\).
Since \(\sqrt{35}\approx{6}\), then \(12-2\sqrt{35}\approx{12-2*6}=0\).
Answer: A
If \(x=(\sqrt{5}-\sqrt{7})^2\), then the best approximation of \(x\) is:
A. \(0\)
B. \(1\)
C. \(2\)
D. \(3\)
E. \(4\)
\(x=(\sqrt{5}-\sqrt{7})^2=5-2\sqrt{35}+7=12-2\sqrt{35}\).
Since \(\sqrt{35}\approx{6}\), then \(12-2\sqrt{35}\approx{12-2*6}=0\).
Answer: A
06 Apr 2015, 20:38
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Fractionalizing the result shows very clearly that the result is closer to 0 than to 1:
x=(√5−√7)^2 = [(5-7)/(√5+√7)]^2
=(-2)^2 / (√5+√7)]^2
= _4_
12+ √35
Now, √35 is close to 6, but undeniably between 5 and 6.
Even if you use both choices for the denominator's range:
12+2(5) = 22 --> x = 4/22 = 2/11
12+2(6) = 24 --> x = 4/24 = 1/6
1/6 < x < 2/11
With x(max) < 2/11, it is abundantly clear--with no room for error--that x is closer to 0.
Cheers!
x=(√5−√7)^2 = [(5-7)/(√5+√7)]^2
=(-2)^2 / (√5+√7)]^2
= _4_
12+ √35
Now, √35 is close to 6, but undeniably between 5 and 6.
Even if you use both choices for the denominator's range:
12+2(5) = 22 --> x = 4/22 = 2/11
12+2(6) = 24 --> x = 4/24 = 1/6
1/6 < x < 2/11
With x(max) < 2/11, it is abundantly clear--with no room for error--that x is closer to 0.
Cheers!
