Question Number 171869 by Mikenice last updated on 21/Jun/22
$${find}\:{the}\:{square}\:{root}\:{of}: \\ $$$$\sqrt{\frac{\mathrm{7}{x}^{\mathrm{2}} +\mathrm{2}\sqrt{\mathrm{14}}\:{x}+\mathrm{2}}{{x}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}{x}\:+\frac{\mathrm{1}}{\mathrm{6}}}} \\ $$
Answered by floor(10²Eta[1]) last updated on 21/Jun/22
$$\mathrm{found}\:\mathrm{it},\:\mathrm{it}'\mathrm{s}\:\mathrm{right}\:\mathrm{above}\:\mathrm{my}\:\mathrm{answer} \\ $$
Commented by Mikenice last updated on 21/Jun/22
$${okay}\:{sir},\:{show}\:{the}\:{solution}\:{sir} \\ $$
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
$$\mathrm{7x}^{\mathrm{2}} +\mathrm{2}\sqrt{\mathrm{14}}\mathrm{x}+\mathrm{2}=\left(\sqrt{\mathrm{7}}\mathrm{x}+\sqrt{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\mathrm{x}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}+\frac{\mathrm{1}}{\mathrm{6}}=\left(\mathrm{x}−\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{16}}+\frac{\mathrm{1}}{\mathrm{6}}=\left(\mathrm{x}−\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{2}} +\frac{\mathrm{5}}{\mathrm{48}} \\ $$$$\sqrt{\frac{\mathrm{7x}^{\mathrm{2}} +\mathrm{2}\sqrt{\mathrm{14}}\mathrm{x}+\mathrm{2}}{\mathrm{x}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}+\frac{\mathrm{1}}{\mathrm{6}}}}=\frac{\mid\sqrt{\mathrm{7}}\mathrm{x}+\sqrt{\mathrm{2}}\mid}{\:\sqrt{\left(\mathrm{x}−\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{2}} +\frac{\mathrm{5}}{\mathrm{48}}}} \\ $$
Commented by mr W last updated on 21/Jun/22
$${maybe}\:{he}\:{meant}\:{actually} \\ $$$$\sqrt{\frac{\mathrm{7x}^{\mathrm{2}} +\mathrm{2}\sqrt{\mathrm{14}}\mathrm{x}+\mathrm{2}}{\mathrm{x}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}+\frac{\mathrm{1}}{\mathrm{16}}}}. \\ $$$${otherwise}\:{it}'{s}\:{a}\:{non}−{sense}\:{question}! \\ $$