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If \( \Large x=3+\sqrt[2]{2} \), then the value of \( \Large \left(\sqrt{x}-\frac{1}{\sqrt{x}}\right) \) is:
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A) 1 |
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B) 2 |
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C) \( \Large 2\sqrt{2} \) |
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D) \( \Large 3\sqrt{3} \) |
Correct Answer:
| B) 2 |
Description for Correct answer:
\( \Large x = 3 + 2\sqrt{2} \)
\( \Large x = 2 + 1 + 2\sqrt{2} = \left(\sqrt{2}+1\right)^{2} \)
\( \Large \sqrt{x} = \sqrt{2}+1 \)
\( \Large \frac{1}{\sqrt{x}} = \frac{1}{\sqrt{2}+1} \)
=> \( \Large \frac{1}{\sqrt{x}}=\frac{1}{\sqrt{2}+1} \times \frac{\sqrt{2}-1}{\sqrt{2}-1} \)
= \( \Large \frac{\sqrt{2}-1}{1} = \sqrt{2}-1 \)
Therefore, \( \Large \sqrt{x} - \frac{1}{\sqrt{x}} \)
= \( \Large \sqrt{2}+1- \left(\sqrt{2}-1\right) \)
= \( \Large \sqrt{2} + 1 - \sqrt{2} + 1 = 2 \)
\( \Large x = 3 + 2\sqrt{2} \)
\( \Large x = 2 + 1 + 2\sqrt{2} = \left(\sqrt{2}+1\right)^{2} \)
\( \Large \sqrt{x} = \sqrt{2}+1 \)
\( \Large \frac{1}{\sqrt{x}} = \frac{1}{\sqrt{2}+1} \)
=> \( \Large \frac{1}{\sqrt{x}}=\frac{1}{\sqrt{2}+1} \times \frac{\sqrt{2}-1}{\sqrt{2}-1} \)
= \( \Large \frac{\sqrt{2}-1}{1} = \sqrt{2}-1 \)
Therefore, \( \Large \sqrt{x} - \frac{1}{\sqrt{x}} \)
= \( \Large \sqrt{2}+1- \left(\sqrt{2}-1\right) \)
= \( \Large \sqrt{2} + 1 - \sqrt{2} + 1 = 2 \)
Part of solved Elementary Mathematics questions and answers : >> Elementary Mathematics
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