Question Number 194226 by BaliramKumar last updated on 30/Jun/23
$$ \\ $$$$\mathrm{If}\:{x}^{\mathrm{2}} \:−\:\mathrm{65}{x}\:=\:\mathrm{64}\sqrt{{x}}\:\mathrm{then}\:\sqrt{{x}\:−\:\sqrt{{x}}\:}\:=\:? \\ $$
Answered by Frix last updated on 30/Jun/23
$${x}^{\mathrm{2}} −\mathrm{65}{x}=\mathrm{64}\sqrt{{x}} \\ $$$${x}=\mathrm{0}\:\Rightarrow\:\sqrt{{x}−\sqrt{{x}}}=\mathrm{0}\:\bigstar \\ $$$${x}^{\frac{\mathrm{3}}{\mathrm{2}}} −\mathrm{65}\sqrt{{x}}−\mathrm{64}=\mathrm{0} \\ $$$${x}={t}^{\mathrm{2}} \wedge{t}>\mathrm{0} \\ $$$${t}^{\mathrm{3}} −\mathrm{65}{t}−\mathrm{64}=\mathrm{0} \\ $$$$\left({t}+\mathrm{1}\right)\left({t}^{\mathrm{2}} −{t}−\mathrm{64}\right)=\mathrm{0} \\ $$$${t}=\frac{\mathrm{1}+\sqrt{\mathrm{257}}}{\mathrm{2}}=\sqrt{{x}} \\ $$$${x}=\frac{\mathrm{129}+\sqrt{\mathrm{257}}}{\mathrm{2}} \\ $$$$\Rightarrow\:\sqrt{{x}−\sqrt{{x}}}=\mathrm{8}\:\bigstar \\ $$
Answered by Frix last updated on 30/Jun/23
$${x}^{\mathrm{2}} −\mathrm{65}{x}−\mathrm{64}\sqrt{{x}}=\mathrm{0} \\ $$$$\sqrt{{x}}\left(\sqrt{{x}}+\mathrm{1}\right)\left({x}−\sqrt{{x}}−\mathrm{64}\right)=\mathrm{0} \\ $$$${x}=\mathrm{0}\:\bigstar \\ $$$$\sqrt{{x}−\sqrt{{x}}}=\mathrm{0} \\ $$$${x}−\sqrt{{x}}−\mathrm{64}=\mathrm{0} \\ $$$${x}−\sqrt{{x}}=\mathrm{64} \\ $$$$\sqrt{{x}−\sqrt{{x}}}=\mathrm{8}\:\bigstar \\ $$
Answered by manxsol last updated on 30/Jun/23
$${x}^{\mathrm{2}} −{x}=\mathrm{64}\left({x}+\sqrt{{x}}\right) \\ $$$$\left(\sqrt{{x}}\right)^{\mathrm{4}} −\left(\sqrt{{x}}\right)^{\mathrm{2}} =\mathrm{64}\left({x}+\sqrt{{x}}\right) \\ $$$$\left(\sqrt{{x}}\:^{\mathrm{2}} −\sqrt{{x}}\right)\left(\sqrt{{x}}\:^{\mathrm{2}} +\sqrt{{x}}\right)=\mathrm{64}\left({x}+\sqrt{{x}}\right) \\ $$$$\left({x}−\sqrt{{x}}\right)\left({x}+\sqrt{{x}}\right)=\mathrm{64}\left({x}+\sqrt{{x}}\right) \\ $$$${x}−\sqrt{{x}}=\mathrm{64} \\ $$$$\sqrt{{x}−\sqrt{{x}}}=\mathrm{8}\:{sol}\:{I} \\ $$$${x}+\sqrt{{x}}=\mathrm{0} \\ $$$$\sqrt{{x}}\left(\sqrt{{x}}+\mathrm{1}\right)=\mathrm{0} \\ $$$${x}=\mathrm{0}\:\:\:\sqrt{{x}−\sqrt{{x}}}\:=\mathrm{0}\:{sol}\:{II} \\ $$$$\:\sqrt{{x}}=−\mathrm{1}\:× \\ $$$$ \\ $$