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555-605 (Medium)|   Algebra|   Roots|      
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guddo
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If x > 0, then 1\((2x)^(1/2) + x^(1/2)) = 25 Jan 2024, 12:17
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Question Stats:

82% (01:37) correct 18% (02:04) wrong based on 540 sessions

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If x > 0, then \(\frac{1}{\sqrt{2x} + \sqrt{x}} = \)

A. \(\frac{1}{\sqrt{3x}}\)

B. \(\frac{1}{2\sqrt{2x}}\)

C. \(\frac{1}{x\sqrt{2}}\)

D. \(\frac{\sqrt{2} - 1}{\sqrt{x}}\)

E. \(\frac{1 + \sqrt{2}}{\sqrt{x}}\)


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If x > 0, then 1\((2x)^(1/2) + x^(1/2)) = 25 Jan 2024, 12:26
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guddo
If x > 0, then \(\frac{1}{\sqrt{2x} + \sqrt{x}} = \)

A. \(\frac{1}{\sqrt{3x}}\)

B. \(\frac{1}{2\sqrt{2x}}\)

C. \(\frac{1}{x\sqrt{2}}\)

D. \(\frac{\sqrt{2} - 1}{\sqrt{x}}\)

E. \(\frac{1 + \sqrt{2}}{\sqrt{x}}\)


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Given:

    \(\frac{1}{\sqrt{2x} + \sqrt{x}} \)

Factor out \(\sqrt{x}\) from the denominator:

    \(\frac{1}{\sqrt{x}(\sqrt{2} + 1)}\)

Rationalize by multiplying the numerator and the denominator by \(\sqrt{2} - 1\):

    \(\frac{\sqrt{2} - 1}{\sqrt{x}(\sqrt{2} + 1)(\sqrt{2} - 1)} = \)

    \(=\frac{\sqrt{2} - 1}{\sqrt{x}(2 - 1)} = \)

    \(=\frac{\sqrt{2} - 1}{\sqrt{x}}\)

Answer: D.
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If x > 0, then 1\((2x)^(1/2) + x^(1/2)) = 25 Jan 2024, 21:57
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guddo
If x > 0, then \(\frac{1}{\sqrt{2x} + \sqrt{x}} = \)

A. \(\frac{1}{\sqrt{3x}}\)

B. \(\frac{1}{2\sqrt{2x}}\)

C. \(\frac{1}{x\sqrt{2}}\)

D. \(\frac{\sqrt{2} - 1}{\sqrt{x}}\)

E. \(\frac{1 + \sqrt{2}}{\sqrt{x}}\)


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If x > 0, then 1\((2x)^(1/2) + x^(1/2)) = 27 Jan 2024, 12:29
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guddo
If x > 0, then \(\frac{1}{\sqrt{2x} + \sqrt{x}} = \)

A. \(\frac{1}{\sqrt{3x}}\)

B. \(\frac{1}{2\sqrt{2x}}\)

C. \(\frac{1}{x\sqrt{2}}\)

D. \(\frac{\sqrt{2} - 1}{\sqrt{x}}\)

E. \(\frac{1 + \sqrt{2}}{\sqrt{x}}\)


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Assume x = 1

\(\frac{1}{\sqrt{2x} + \sqrt{x}} = \frac{1}{\sqrt{2*1} + \sqrt{1}} = \frac{1}{\sqrt{2} + \sqrt{1}}\)

Multiplying numerator and denominator by (\(\sqrt{2}-\sqrt{1}\))

\(\frac{1* (\sqrt{2}-\sqrt{1})}{(\sqrt{2} + \sqrt{1})* (\sqrt{2}-\sqrt{1})}\)

\(\frac{\sqrt{2}-\sqrt{1}}{2-1}\)

\(\sqrt{2}-\sqrt{1}\)

Answer choice elimination

A. \(\frac{1}{\sqrt{3x}}\)

x = 1 ⇒

\(\frac{1}{\sqrt{3}}\) ⇒ Eliminate

B. \(\frac{1}{2\sqrt{2x}}\)

x = 1 ⇒

\(\frac{1}{\sqrt{2\sqrt{2}}}\) ⇒ Eliminate

C. \(\frac{1}{x\sqrt{2}}\)

x = 1 ⇒

\(\frac{1}{\sqrt{2}}\) ⇒ Eliminate

D. \(\frac{\sqrt{2} - 1}{\sqrt{x}}\)

x = 1 ⇒

\(\frac{\sqrt{2} - 1}{\sqrt{1}}\) ⇒ Keep

E. \(\frac{1 + \sqrt{2}}{\sqrt{x}}\)

x = 1 ⇒

\(\frac{1 + \sqrt{2}}{\sqrt{1}}\) ⇒ Eliminate
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If x > 0, then 1\((2x)^(1/2) + x^(1/2)) = 07 Apr 2026, 12:34
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