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4(\sqrt6 + \sqrt2)
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16 Apr 2019, 02:03
16 Apr 2019, 02:03
Expert Reply
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Question Stats:
73% (02:32) correct
26% (02:51) wrong based on 41 sessions
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\(\frac{4(\sqrt{6} + \sqrt{2})}{\sqrt{6}-\sqrt{2}} - \frac{2+\sqrt{3}}{2-\sqrt{3}} =\)
A. \(1\)
B. \(\sqrt{6} - \sqrt{2}\)
C. \(\sqrt{6} + \sqrt{2}\)
D. \(8\)
E. \(12\)
A. \(1\)
B. \(\sqrt{6} - \sqrt{2}\)
C. \(\sqrt{6} + \sqrt{2}\)
D. \(8\)
E. \(12\)
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Re: 4(\sqrt6 + \sqrt2)
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26 Apr 2020, 14:18
26 Apr 2020, 14:18
2
1
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Carcass wrote:
\(\frac{4(\sqrt{6} + \sqrt{2})}{\sqrt{6}-\sqrt{2}} - \frac{2+\sqrt{3}}{2-\sqrt{3}} =\)
A. \(1\)
B. \(\sqrt{6} - \sqrt{2}\)
C. \(\sqrt{6} + \sqrt{2}\)
D. \(8\)
E. \(12\)
A. \(1\)
B. \(\sqrt{6} - \sqrt{2}\)
C. \(\sqrt{6} + \sqrt{2}\)
D. \(8\)
E. \(12\)
\(\frac{4(\sqrt{6} + \sqrt{2})}{\sqrt{6}-\sqrt{2}} - \frac{2+\sqrt{3}}{2-\sqrt{3}} =\frac{4(\sqrt{6} + \sqrt{2})(\sqrt{6}+\sqrt{2})}{(\sqrt{6}-\sqrt{2})(\sqrt{6}+\sqrt{2})} - \frac{(2+\sqrt{3})(2+\sqrt{3})}{(2-\sqrt{3})(2+\sqrt{3})}\)
\(=\frac{4(6 + 2\sqrt{12} + 2)}{6-2} - \frac{4+4\sqrt{3}+3}{4-3}\)
\(=\frac{4(8 + 2\sqrt{12})}{4} - \frac{7+4\sqrt{3}}{1}\)
\(=8 + 2\sqrt{12} - (7+4\sqrt{3})\)
\(=8 + 4\sqrt{3} - 7-4\sqrt{3}\)
\(= 1\)
Answer: A
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Brent
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Re: 4(\sqrt6 + \sqrt2)
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12 May 2020, 23:09
12 May 2020, 23:09
1
Carcass wrote:
\(\frac{4(\sqrt{6} + \sqrt{2})}{\sqrt{6}-\sqrt{2}} - \frac{2+\sqrt{3}}{2-\sqrt{3}} =\)
A. \(1\)
B. \(\sqrt{6} - \sqrt{2}\)
C. \(\sqrt{6} + \sqrt{2}\)
D. \(8\)
E. \(12\)
A. \(1\)
B. \(\sqrt{6} - \sqrt{2}\)
C. \(\sqrt{6} + \sqrt{2}\)
D. \(8\)
E. \(12\)
We can rationalize each term as follows:
4(√6 + √2)/(√6 - √2) = 4(√6 + √2)(√6 + √2)/(√6 - √2)(√6 + √2)
= 4[6 + 2 + 2√6√2]/[6 - 2]
= 8 + 4√3
(2 + √3)/(2 - √3) = (2 + √3)(2 + √3)/(2 - √3)(2 + √3)
= [4 + 3 + 2(2)(√3)]/[4 - 3]
= 7 + 4√3
Thus, subtracting, we have: 1
Answer A
