x128−−−√=162−−−√xx128=162x\frac{x}{{\sqrt {128} }} = \frac{{\sqrt {162} }}{x}" /> x128−−−√=162−−−√xx128=162x\frac{x}{{\sqrt {128} }} = \frac{{\sqrt {162} }}{x}" /> x128−−−√=162−−−√xx128=162x\frac{x}{{\sqrt {128} }} = \frac{{\sqrt {162} }}{x}" />

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What should come in place of both x in the equation $\frac{x}{\sqrt{128}} = \frac{\sqrt{162}}{x}$ $\frac{x}{{\sqrt {128} }} = \frac{{\sqrt {162} }}{x}$

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Answer:- 1

Explanation:-

Solution:

\begin{aligned} Let \frac{x}{\sqrt{128}} = \frac{\sqrt{162}}{x} \\ Then x^{2} = \sqrt{128 \times 162} \\ = \sqrt{64 \times 2 \times 18 \times 9} \\ = \sqrt{8^{2} \times 6^{2} \times 3^{2}} \\ = 8 \times 6 \times 3 \\ = 144 \\ ∴ x = \sqrt{144} = 12 \end{aligned}

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", "text": "What should come in place of both x in the equation

\frac{x}{\sqrt{128}} = \frac{\sqrt{162}}{x}

$\frac{x}{{\sqrt {128} }} = \frac{{\sqrt {162} }}{x}$ ", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" }, "answerCount": "4", "acceptedAnswer": { "@type": "Answer", "text": "

Solution:

\begin{aligned} Let \frac{x}{\sqrt{128}} = \frac{\sqrt{162}}{x} \\ Then x^{2} = \sqrt{128 \times 162} \\ = \sqrt{64 \times 2 \times 18 \times 9} \\ = \sqrt{8^{2} \times 6^{2} \times 3^{2}} \\ = 8 \times 6 \times 3 \\ = 144 \\ ∴ x = \sqrt{144} = 12 \end{aligned}

$\eqalign{ & {\text{Let}}\,\frac{x}{{\sqrt {128} }} = \frac{{\sqrt {162} }}{x} \cr & {\text{Then}}\,{x^2} = \sqrt {128 \times 162} \cr & = \sqrt {64 \times 2 \times 18 \times 9} \cr & = \sqrt {{8^2} \times {6^2} \times {3^2}} \cr & = 8 \times 6 \times 3 \cr & = 144 \cr & \therefore x = \sqrt {144} = 12 \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" } }, "suggestedAnswer": { "@type": "Answer", "text": "

Solution: