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11 Jan 2022, 04:30
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Difficulty:
75%
(hard)
Question Stats:
58% (02:19) correct
42%
(02:43)
wrong
based on 161
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History
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Time
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If \(x = 2 +\sqrt{3}\), then the value of \(\sqrt{x}+\frac{1}{\sqrt{x}}\) is equal to
A. \(3 + \sqrt{3}\)
B. \(\sqrt{6}\)
C. \(2\sqrt{6}\)
D. \(6\)
E. \(3\sqrt{3}\)
Are You Up For the Challenge: 700 Level Questions
A. \(3 + \sqrt{3}\)
B. \(\sqrt{6}\)
C. \(2\sqrt{6}\)
D. \(6\)
E. \(3\sqrt{3}\)
Are You Up For the Challenge: 700 Level Questions
12 Jan 2022, 05:19
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Square \(\sqrt{x} + \frac{1}{\sqrt{x}}\) to get
\((\sqrt{x} + \frac{1}{\sqrt{x}})^2\)
\(= x + 2 + \frac{1}{x}\)
Sub in \(x = 2 + \sqrt{3}\) to get
\(2 + \sqrt{3} + 2 + \frac{1}{2+\sqrt{3}}\)
\(= 4 + \sqrt{3} + \frac{1}{2+\sqrt{3}}\)
\(=\frac{8+4\sqrt{3} + 2\sqrt{3}+3+1}{2+\sqrt{3}}\)
\(= \frac{12 + 6\sqrt{3}}{2+\sqrt{3}}\)
\(= \frac{6(2+\sqrt{3})}{2+\sqrt{3}}\)
\(= 6\)
Because of the squaring done earlier, the square root of the answer must be taken
\(\sqrt{6}\) Answer B
\((\sqrt{x} + \frac{1}{\sqrt{x}})^2\)
\(= x + 2 + \frac{1}{x}\)
Sub in \(x = 2 + \sqrt{3}\) to get
\(2 + \sqrt{3} + 2 + \frac{1}{2+\sqrt{3}}\)
\(= 4 + \sqrt{3} + \frac{1}{2+\sqrt{3}}\)
\(=\frac{8+4\sqrt{3} + 2\sqrt{3}+3+1}{2+\sqrt{3}}\)
\(= \frac{12 + 6\sqrt{3}}{2+\sqrt{3}}\)
\(= \frac{6(2+\sqrt{3})}{2+\sqrt{3}}\)
\(= 6\)
Because of the squaring done earlier, the square root of the answer must be taken
\(\sqrt{6}\) Answer B
10 Aug 2022, 07:40
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[quote="Bunuel"]If \(x = 2 +\sqrt{3}\), then the value of \(\sqrt{x}+\frac{1}{\sqrt{x}}\) is equal to
A. \(3 + \sqrt{3}\)
B. \(\sqrt{6}\)
C. \(2\sqrt{6}\)
D. \(6\)
E. \(3\sqrt{3}\)
Can anyone explain how 1/sqrt(x) * sqrt(x) becomes 2 when this is "foiled out"? I am not able to arrive at x+2+(1/x) and it is the "2" that I cannot get, but I understand the rest.
A. \(3 + \sqrt{3}\)
B. \(\sqrt{6}\)
C. \(2\sqrt{6}\)
D. \(6\)
E. \(3\sqrt{3}\)
Can anyone explain how 1/sqrt(x) * sqrt(x) becomes 2 when this is "foiled out"? I am not able to arrive at x+2+(1/x) and it is the "2" that I cannot get, but I understand the rest.
