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If \( \Large x=7-4\sqrt{3} \), then \( \Large \sqrt{x}+\frac{1}{\sqrt{x}} \) is equal to:
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A) 1 |
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B) 2 |
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C) 3 |
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D) 4 |
Correct Answer:
| D) 4 |
Description for Correct answer:
\( \Large x = 7 - 4\sqrt{3} \)
= \( \Large 4 + 3 - 4\sqrt{3} \)
= \( \Large \left(2\right)^{2}+ \left(\sqrt{3}\right)^{2}-2 \times 2\sqrt{3} \)
= \( \Large \left(2-\sqrt{3}\right)^{2} \)
Therefore, \( \Large \left[ a^{2}+b^{2}-2ab= \left(a-b\right)^{2} \right] \)
=> \( \Large x = \left(2-\sqrt{3}\right)^{2} \)
\( \Large \sqrt{x} = 2-\sqrt{3} \)
\( \Large \frac{1}{\sqrt{x}}=\frac{1}{2-\sqrt{3}} \times \frac{2+\sqrt{3}}{2+\sqrt{3}} \)
= \( \Large 2+\sqrt{3} \)
Therefore, \( \Large \sqrt{x}+\frac{1}{\sqrt{x}} \)
\( \Large 2-\sqrt{3}+2+\sqrt{3}=4 \)
\( \Large x = 7 - 4\sqrt{3} \)
= \( \Large 4 + 3 - 4\sqrt{3} \)
= \( \Large \left(2\right)^{2}+ \left(\sqrt{3}\right)^{2}-2 \times 2\sqrt{3} \)
= \( \Large \left(2-\sqrt{3}\right)^{2} \)
Therefore, \( \Large \left[ a^{2}+b^{2}-2ab= \left(a-b\right)^{2} \right] \)
=> \( \Large x = \left(2-\sqrt{3}\right)^{2} \)
\( \Large \sqrt{x} = 2-\sqrt{3} \)
\( \Large \frac{1}{\sqrt{x}}=\frac{1}{2-\sqrt{3}} \times \frac{2+\sqrt{3}}{2+\sqrt{3}} \)
= \( \Large 2+\sqrt{3} \)
Therefore, \( \Large \sqrt{x}+\frac{1}{\sqrt{x}} \)
\( \Large 2-\sqrt{3}+2+\sqrt{3}=4 \)
Part of solved Elementary Mathematics questions and answers : >> Elementary Mathematics
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