Question Number 167772 by mathlove last updated on 24/Mar/22
$${x}+\sqrt{{x}}=\mathrm{5} \\ $$$${x}+\frac{\mathrm{5}}{\:\sqrt{{x}}}=? \\ $$
Commented by mr W last updated on 24/Mar/22
$$\left(\sqrt{{x}}\right)^{\mathrm{2}} +\sqrt{{x}}−\mathrm{5}=\mathrm{0} \\ $$$$\sqrt{{x}}=\frac{\sqrt{\mathrm{21}}−\mathrm{1}}{\mathrm{2}} \\ $$$${x}+\frac{\mathrm{5}}{\:\sqrt{{x}}}=\mathrm{5}+\frac{\mathrm{5}}{\:\sqrt{{x}}}−\sqrt{{x}} \\ $$$$=\mathrm{5}+\frac{\mathrm{10}}{\:\sqrt{\mathrm{21}}−\mathrm{1}}−\frac{\sqrt{\mathrm{21}}−\mathrm{1}}{\mathrm{2}} \\ $$$$=\mathrm{5}+\frac{\sqrt{\mathrm{21}}+\mathrm{1}}{\:\mathrm{2}}−\frac{\sqrt{\mathrm{21}}−\mathrm{1}}{\mathrm{2}} \\ $$$$=\mathrm{6} \\ $$
Commented by MJS_new last updated on 24/Mar/22
$${x}+\sqrt{{x}}={a}\:\Rightarrow\:{x}+\frac{{a}}{\:\sqrt{{x}}}={a}+\mathrm{1} \\ $$$$\mathrm{prove}: \\ $$$$\left(\mathrm{1}\right) \\ $$$${a}={x}+\sqrt{{x}}=\left(\sqrt{{x}}+\mathrm{1}\right)\sqrt{{x}}=\frac{\left(\mathrm{1}+\sqrt{{x}}\right)\left(\mathrm{1}−\sqrt{{x}}\right)\sqrt{{x}}}{\mathrm{1}−\sqrt{{x}}}= \\ $$$$=\frac{\left(\mathrm{1}−{x}\right)\sqrt{{x}}}{\mathrm{1}−\sqrt{{x}}} \\ $$$$\left(\mathrm{2}\right) \\ $$$${a}=\frac{\left(\mathrm{1}−{x}\right)\sqrt{{x}}}{\mathrm{1}−\sqrt{{x}}} \\ $$$${a}\left(\mathrm{1}−\sqrt{{x}}\right)=\left(\mathrm{1}−{x}\right)\sqrt{{x}} \\ $$$${a}−{a}\sqrt{{x}}=\left(\mathrm{1}−{x}\right)\sqrt{{x}} \\ $$$$\frac{{a}}{\:\sqrt{{x}}}−{a}=\mathrm{1}−{x} \\ $$$${x}−\frac{{a}}{\:\sqrt{{x}}}={a}+\mathrm{1} \\ $$
Answered by mnjuly1970 last updated on 25/Mar/22
$$\:\:{x}\:+\:\frac{{x}+\sqrt{{x}}}{\:\sqrt{{x}}}=\:{x}+\sqrt{{x}}\:+\mathrm{1}=\:\mathrm{6} \\ $$