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Question Number 171869 by Mikenice last updated on 21/Jun/22

$$findthesquarerootof:$$$$\sqrt{\frac{7x^2+2\sqrt{14}x+2}{x^2-\frac{1}{2}x+\frac{1}{6}}}$$

Answered by floor(10²Eta[1]) last updated on 21/Jun/22

$$\mathrm{found}\mathrm{it},{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{right}\mathrm{above}\mathrm{my}\mathrm{answer}$$

Commented by Mikenice last updated on 21/Jun/22

$$okaysir,showthesolutionsir$$

Commented by floor(10²Eta[1]) last updated on 21/Jun/22

$$\mathrm{what}\mathrm{solution}????\mathrm{do}\mathrm{you}\mathrm{want}\mathrm{that}$$$$\mathrm{expression}\mathrm{a}\mathrm{simplest}\mathrm{form}??$$

Commented by Mikenice last updated on 21/Jun/22

$$yes$$

Answered by floor(10²Eta[1]) last updated on 21/Jun/22

$${7\mathrm{x}}^2+2\sqrt{14}\mathrm{x}+2={(\sqrt{7}\mathrm{x}+\sqrt{2})}^2$$$${\mathrm{x}}^2-\frac{1}{2}\mathrm{x}+\frac{1}{6}={(\mathrm{x}-\frac{1}{4})}^2-\frac{1}{16}+\frac{1}{6}={(\mathrm{x}-\frac{1}{4})}^2+\frac{5}{48}$$$$\sqrt{\frac{{7\mathrm{x}}^2+2\sqrt{14}\mathrm{x}+2}{{\mathrm{x}}^2-\frac{1}{2}\mathrm{x}+\frac{1}{6}}}=\frac{\mid \sqrt{7}\mathrm{x}+\sqrt{2}\mid }{\sqrt{{(\mathrm{x}-\frac{1}{4})}^2+\frac{5}{48}}}$$

Commented by mr W last updated on 21/Jun/22

$$maybehemeantactually$$$$\sqrt{\frac{{7\mathrm{x}}^2+2\sqrt{14}\mathrm{x}+2}{{\mathrm{x}}^2-\frac{1}{2}\mathrm{x}+\frac{1}{16}}}.$$$$otherwise{it}^{\prime }sanon-sensequestion!$$

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