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Question Number 53491 by ajfour last updated on 22/Jan/19

(((1+x)/(√x)))^2 +2a(((1+x)/(√x)))+1=0 solve for x.

$$\left(\frac{\mathrm{1}+{x}}{\sqrt{{x}}}\right)^{\mathrm{2}} +\mathrm{2}{a}\left(\frac{\mathrm{1}+{x}}{\sqrt{{x}}}\right)+\mathrm{1}=\mathrm{0} \\ $$$${solve}\:{for}\:{x}. \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 22/Jan/19

k^2 +2ak+a^2 +1=a^2 (k+a)^2 =a^2 −1 k=−a±(√(a^2 −1)) ((1+x)/(√x))=−a±(√(a^2 −1)) 1+x=((√x)/b) [(1/b)=−a±(√(a^2 −1)) ] x=b^2 x^2 +2b^2 x+b^2 x^2 (b^2 )+x(2b^2 −1)+b^2 =0 x=((−(2b^2 −1)±(√((2b^2 −1)^2 −4b^2 ×b^2 )) )/(2b^2 )) x=((−(2b^2 −1)±(√(4b^4 −4b^2 +1−4b^4 )))/(2b^2 )) x=((−(2b^2 −1)±(√(1−4b^2 )))/(2b^2 )) pls subdtitude the value of b...

$${k}^{\mathrm{2}} +\mathrm{2}{ak}+{a}^{\mathrm{2}} +\mathrm{1}={a}^{\mathrm{2}} \\ $$$$\left({k}+{a}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} −\mathrm{1} \\ $$$${k}=−{a}\pm\sqrt{{a}^{\mathrm{2}} −\mathrm{1}}\: \\ $$$$\frac{\mathrm{1}+{x}}{\sqrt{{x}}}=−{a}\pm\sqrt{{a}^{\mathrm{2}} −\mathrm{1}}\: \\ $$$$\mathrm{1}+{x}=\frac{\sqrt{{x}}}{{b}}\:\:\:\:\left[\frac{\mathrm{1}}{{b}}=−{a}\pm\sqrt{{a}^{\mathrm{2}} −\mathrm{1}}\:\right] \\ $$$${x}={b}^{\mathrm{2}} {x}^{\mathrm{2}} +\mathrm{2}{b}^{\mathrm{2}} {x}+{b}^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} \left({b}^{\mathrm{2}} \right)+{x}\left(\mathrm{2}{b}^{\mathrm{2}} −\mathrm{1}\right)+{b}^{\mathrm{2}} =\mathrm{0} \\ $$$${x}=\frac{−\left(\mathrm{2}{b}^{\mathrm{2}} −\mathrm{1}\right)\pm\sqrt{\left(\mathrm{2}{b}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} −\mathrm{4}{b}^{\mathrm{2}} ×{b}^{\mathrm{2}} }\:}{\mathrm{2}{b}^{\mathrm{2}} } \\ $$$${x}=\frac{−\left(\mathrm{2}{b}^{\mathrm{2}} −\mathrm{1}\right)\pm\sqrt{\mathrm{4}{b}^{\mathrm{4}} −\mathrm{4}{b}^{\mathrm{2}} +\mathrm{1}−\mathrm{4}{b}^{\mathrm{4}} }}{\mathrm{2}{b}^{\mathrm{2}} } \\ $$$${x}=\frac{−\left(\mathrm{2}{b}^{\mathrm{2}} −\mathrm{1}\right)\pm\sqrt{\mathrm{1}−\mathrm{4}{b}^{\mathrm{2}} }}{\mathrm{2}{b}^{\mathrm{2}} } \\ $$$${pls}\:{subdtitude}\:{the}\:{value}\:{of}\:{b}... \\ $$

Answered by malwaan last updated on 22/Jan/19

((1+x)/(√x))=((−2a±(√(4a^2 −4)))/2) =−a±(√(a^2 −1)) ∴((1+2x+x^2 )/x)=a^2 ±2a(√(a^2 −1))+a^2 −1 ((1+x^2 )/x)+2=2a^2 ±2a(√(a^2 −1))−1 1+x^2 =(2a^2 ±2a(√(a^2 −1))−3)x x^2 −(2a^2 ±2a(√(a^2 −1))−3)x+1=0 x=((2a^2 ±2a(√(a^2 −1))−3±(√((2a^2 ±2a(√(a^2 −1))−3)^2 −4)))/2) Am I right ?

$$\frac{\mathrm{1}+{x}}{\sqrt{{x}}}=\frac{−\mathrm{2}{a}\pm\sqrt{\mathrm{4}{a}^{\mathrm{2}} −\mathrm{4}}}{\mathrm{2}} \\ $$$$=−{a}\pm\sqrt{{a}^{\mathrm{2}} −\mathrm{1}} \\ $$$$\therefore\frac{\mathrm{1}+\mathrm{2}{x}+{x}^{\mathrm{2}} }{{x}}={a}^{\mathrm{2}} \pm\mathrm{2}{a}\sqrt{{a}^{\mathrm{2}} −\mathrm{1}}+{a}^{\mathrm{2}} −\mathrm{1} \\ $$$$\frac{\mathrm{1}+{x}^{\mathrm{2}} }{{x}}+\mathrm{2}=\mathrm{2}{a}^{\mathrm{2}} \pm\mathrm{2}{a}\sqrt{{a}^{\mathrm{2}} −\mathrm{1}}−\mathrm{1} \\ $$$$\mathrm{1}+{x}^{\mathrm{2}} =\left(\mathrm{2}{a}^{\mathrm{2}} \pm\mathrm{2}{a}\sqrt{{a}^{\mathrm{2}} −\mathrm{1}}−\mathrm{3}\right){x} \\ $$$${x}^{\mathrm{2}} −\left(\mathrm{2}{a}^{\mathrm{2}} \pm\mathrm{2}{a}\sqrt{{a}^{\mathrm{2}} −\mathrm{1}}−\mathrm{3}\right){x}+\mathrm{1}=\mathrm{0} \\ $$$${x}=\frac{\mathrm{2}{a}^{\mathrm{2}} \pm\mathrm{2}{a}\sqrt{{a}^{\mathrm{2}} −\mathrm{1}}−\mathrm{3}\pm\sqrt{\left(\mathrm{2}{a}^{\mathrm{2}} \pm\mathrm{2}{a}\sqrt{{a}^{\mathrm{2}} −\mathrm{1}}−\mathrm{3}\right)^{\mathrm{2}} −\mathrm{4}}}{\mathrm{2}} \\ $$$$\boldsymbol{{Am}}\:\boldsymbol{{I}}\:\boldsymbol{{right}}\:? \\ $$

Answered by mr W last updated on 22/Jan/19

let u=((1+x)/(√x))=(1/(√x))+(√x)≥2 u^2 +2au+1=0 (u+a)^2 =a^2 −1 u=−a±(√(a^2 −1))≥2 (√(a^2 −1))≥a+2 ⇒a≤−(5/4) x−u(√x)+1=0 (√x)=((u±(√(u^2 −4)))/2) x=((u^2 −2±u(√(u^2 −4)))/2) ⇒x=((2a^2 −3±2a(√(a^2 −1))±(−a+(√(a^2 −1)))(√(2a^2 −5±2a(√(a^2 −1)))))/2)

$${let}\:{u}=\frac{\mathrm{1}+{x}}{\sqrt{{x}}}=\frac{\mathrm{1}}{\sqrt{{x}}}+\sqrt{{x}}\geqslant\mathrm{2} \\ $$$${u}^{\mathrm{2}} +\mathrm{2}{au}+\mathrm{1}=\mathrm{0} \\ $$$$\left({u}+{a}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} −\mathrm{1} \\ $$$${u}=−{a}\pm\sqrt{{a}^{\mathrm{2}} −\mathrm{1}}\geqslant\mathrm{2} \\ $$$$\sqrt{{a}^{\mathrm{2}} −\mathrm{1}}\geqslant{a}+\mathrm{2} \\ $$$$\Rightarrow{a}\leqslant−\frac{\mathrm{5}}{\mathrm{4}} \\ $$$${x}−{u}\sqrt{{x}}+\mathrm{1}=\mathrm{0} \\ $$$$\sqrt{{x}}=\frac{{u}\pm\sqrt{{u}^{\mathrm{2}} −\mathrm{4}}}{\mathrm{2}} \\ $$$${x}=\frac{{u}^{\mathrm{2}} −\mathrm{2}\pm{u}\sqrt{{u}^{\mathrm{2}} −\mathrm{4}}}{\mathrm{2}} \\ $$$$\Rightarrow{x}=\frac{\mathrm{2}{a}^{\mathrm{2}} −\mathrm{3}\pm\mathrm{2}{a}\sqrt{{a}^{\mathrm{2}} −\mathrm{1}}\pm\left(−{a}+\sqrt{{a}^{\mathrm{2}} −\mathrm{1}}\right)\sqrt{\mathrm{2}{a}^{\mathrm{2}} −\mathrm{5}\pm\mathrm{2}{a}\sqrt{{a}^{\mathrm{2}} −\mathrm{1}}}}{\mathrm{2}} \\ $$

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