Question Number 115544 by bemath last updated on 26/Sep/20
$${Given}\:{x}^{\mathrm{2}} +\mathrm{12}\sqrt{{x}}\:=\:\mathrm{5} \\ $$$${then}\:{x}+\mathrm{2}\sqrt{{x}}\:? \\ $$
Commented by MJS_new last updated on 26/Sep/20
$${x}^{\mathrm{2}} +\mathrm{12}\sqrt{{x}}=\mathrm{5} \\ $$$$\mathrm{has}\:\mathrm{got}\:\mathrm{1}\:\mathrm{real}\:\mathrm{and}\:\mathrm{2}\:\mathrm{complex}\:\mathrm{solutions}: \\ $$$${x}=\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}\vee{x}=−\mathrm{3}\pm\mathrm{4i} \\ $$$$\Rightarrow\:{x}+\mathrm{2}\sqrt{{x}}=\mathrm{1}\vee{x}+\mathrm{2}\sqrt{{x}}=−\mathrm{1}\pm\mathrm{8i} \\ $$
Answered by bobhans last updated on 26/Sep/20
$${let}\:\sqrt{{x}}\:=\:{w}\:;\:{w}\geqslant\mathrm{0}\Rightarrow\:{w}^{\mathrm{4}} +\mathrm{12}{w}−\mathrm{5}=\mathrm{0} \\ $$$${factoring} \\ $$$$\left({w}^{\mathrm{2}} +\mathrm{2}{w}−\mathrm{1}\right)\left({w}^{\mathrm{2}} −\mathrm{2}{w}+\mathrm{5}\right)=\mathrm{0} \\ $$$${for}\:{w}^{\mathrm{2}} −\mathrm{2}{w}+\mathrm{5}\:>\:\mathrm{0}\:,\:\forall{w}\in\mathbb{R} \\ $$$${for}\:{w}^{\mathrm{2}} +\mathrm{2}{w}−\mathrm{1}=\mathrm{0} \\ $$$${w}^{\mathrm{2}} +\mathrm{2}{w}\:=\:\mathrm{1}\:\Rightarrow\:\left(\sqrt{{x}}\right)^{\mathrm{2}} +\mathrm{2}\sqrt{{x}}\:=\:\mathrm{1} \\ $$$$\Rightarrow\:{x}+\mathrm{2}\sqrt{{x}}\:=\:\mathrm{1} \\ $$