Question-205334

Question Number 205334 by cortano12 last updated on 17/Mar/24

Commented by Ghisom last updated on 17/Mar/24

x^2 −(5/2)x^(3/2) +2x−(5/4)x^(1/2) +(1/4)=0 (x−2x^(1/2) +(1/2))(x−(1/2)x^(1/2) +(1/2))=0 x=(3/2)±(√2) x=−(3/8)±((√7)/8)i

$${x}^{\mathrm{2}} −\frac{\mathrm{5}}{\mathrm{2}}{x}^{\mathrm{3}/\mathrm{2}} +\mathrm{2}{x}−\frac{\mathrm{5}}{\mathrm{4}}{x}^{\mathrm{1}/\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{4}}=\mathrm{0} \\ $$$$\left({x}−\mathrm{2}{x}^{\mathrm{1}/\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}\right)\left({x}−\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{1}/\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}\right)=\mathrm{0} \\ $$$${x}=\frac{\mathrm{3}}{\mathrm{2}}\pm\sqrt{\mathrm{2}} \\ $$$${x}=−\frac{\mathrm{3}}{\mathrm{8}}\pm\frac{\sqrt{\mathrm{7}}}{\mathrm{8}}\mathrm{i} \\ $$

Answered by Rasheed.Sindhi last updated on 17/Mar/24

10(√x) +(5/( (√x) ))=4x+(1/x)+8 10(√x) +(5/( (√x) ))=(2(√x) )^2 +((1/( (√x) )))^2 +8 5(2(√x) +(1/( (√x) )))=(2(√x) +(1/( (√x) )))^2 +4 y^2 −5y+4=0 ; y=2(√x) +(1/( (√x) )) (y−1)(y−4)=0 y=1 or y=4 2(√x) +(1/( (√x) ))=1 or 2(√x) +(1/( (√x) ))=4 2x−(√x) +1=0 or 2x−4(√x) +1=0 { ((2((√x) )^2 −(√x) +1=0⇒(√x) =((1±i(√7))/4))),((2((√x) )^2 −4(√x) +1=0⇒(√x) =((4±2(√2))/4)=((2±(√2))/2))) :} { ((x=((−3±i(√7) )/8))),((x=((6±4(√2))/4)=((3±2(√2) )/2))) :}

$$\mathrm{10}\sqrt{{x}}\:+\frac{\mathrm{5}}{\:\sqrt{{x}}\:}=\mathrm{4}{x}+\frac{\mathrm{1}}{{x}}+\mathrm{8} \\ $$$$\mathrm{10}\sqrt{{x}}\:+\frac{\mathrm{5}}{\:\sqrt{{x}}\:}=\left(\mathrm{2}\sqrt{{x}}\:\right)^{\mathrm{2}} +\left(\frac{\mathrm{1}}{\:\sqrt{{x}}\:}\right)^{\mathrm{2}} +\mathrm{8} \\ $$$$\mathrm{5}\left(\mathrm{2}\sqrt{{x}}\:+\frac{\mathrm{1}}{\:\sqrt{{x}}\:}\right)=\left(\mathrm{2}\sqrt{{x}}\:+\frac{\mathrm{1}}{\:\sqrt{{x}}\:}\right)^{\mathrm{2}} +\mathrm{4} \\ $$$${y}^{\mathrm{2}} −\mathrm{5}{y}+\mathrm{4}=\mathrm{0}\:\:;\:{y}=\mathrm{2}\sqrt{{x}}\:+\frac{\mathrm{1}}{\:\sqrt{{x}}\:} \\ $$$$\left({y}−\mathrm{1}\right)\left({y}−\mathrm{4}\right)=\mathrm{0} \\ $$$${y}=\mathrm{1}\:\mathrm{or}\:{y}=\mathrm{4} \\ $$$$\mathrm{2}\sqrt{{x}}\:+\frac{\mathrm{1}}{\:\sqrt{{x}}\:}=\mathrm{1}\:\mathrm{or}\:\mathrm{2}\sqrt{{x}}\:+\frac{\mathrm{1}}{\:\sqrt{{x}}\:}=\mathrm{4} \\ $$$$\mathrm{2}{x}−\sqrt{{x}}\:+\mathrm{1}=\mathrm{0}\:\mathrm{or}\:\mathrm{2}{x}−\mathrm{4}\sqrt{{x}}\:+\mathrm{1}=\mathrm{0} \\ $$$$\begin{cases}{\mathrm{2}\left(\sqrt{{x}}\:\right)^{\mathrm{2}} −\sqrt{{x}}\:+\mathrm{1}=\mathrm{0}\Rightarrow\sqrt{{x}}\:=\frac{\mathrm{1}\pm{i}\sqrt{\mathrm{7}}}{\mathrm{4}}}\\{\mathrm{2}\left(\sqrt{{x}}\:\right)^{\mathrm{2}} −\mathrm{4}\sqrt{{x}}\:+\mathrm{1}=\mathrm{0}\Rightarrow\sqrt{{x}}\:=\frac{\mathrm{4}\pm\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{4}}=\frac{\mathrm{2}\pm\sqrt{\mathrm{2}}}{\mathrm{2}}}\end{cases}\: \\ $$$$\begin{cases}{{x}=\frac{−\mathrm{3}\pm{i}\sqrt{\mathrm{7}}\:}{\mathrm{8}}}\\{{x}=\frac{\mathrm{6}\pm\mathrm{4}\sqrt{\mathrm{2}}}{\mathrm{4}}=\frac{\mathrm{3}\pm\mathrm{2}\sqrt{\mathrm{2}}\:}{\mathrm{2}}}\end{cases} \\ $$

Commented by Rasheed.Sindhi last updated on 17/Mar/24

$${Corrected}\:{sir} \\ $$

Commented by cortano12 last updated on 17/Mar/24

$$\:\mathrm{10}\sqrt{\mathrm{x}}\:+\frac{\mathrm{5}}{\:\sqrt{\mathrm{x}}}\:=\:\mathrm{4x}+\frac{\mathrm{1}}{\mathrm{x}}+\mathrm{8}\: \\ $$

Facebook Tweet Pin

Leave a Reply Cancel reply