Question Number 161507 by cortano last updated on 18/Dec/21
Answered by MathsFan last updated on 19/Dec/21
$${x}=\pm\mathrm{0}.\mathrm{51} \\ $$
Answered by mr W last updated on 19/Dec/21
)=x^2 (x(√(x+a))−1) (x(√(x+a))−1)(x(√(x+a))+1)(√(x+a))=x^2 (x(√(x+a))−1) (x(√(x+a))−1)(ax+(√(x+a)))=0 case 1: x(√(x+a))−1=0 (√(x+a))=(1/x) ⇒x>0 (1/x^2 )=x+a (1/(x^3 ))−(a/x)−1=0 ⇒(1/x)=2(√(a/3)) sin (−(1/3)sin^(−1) ((3(√3))/(2a(√a))) +((2kπ)/3)) with k=0,1,2 since x>0, ⇒x=((√3)/(2(√((√5)+1)) sin (((2π)/3)−(1/3) sin^(−1) ((3(√3))/(2((√5)+1)(√((√5)+1)))))))≈0.516239 case 2: ax+(√(x+a))=0 (√(x+a))=−ax ⇒x<0 a^2 x^2 −x−a=0 ⇒x=((1−(√(1+4a^3 )))/(2a^2 ))=((1−(√(65+32(√5))))/(4(3+(√5))))≈−0.510194](Q161532.png)
$${let}\:{a}=\sqrt{\mathrm{5}}+\mathrm{1}\:{just}\:{for}\:{easy}\:{writing} \\ $$$$\frac{\mathrm{1}}{\:\sqrt{{x}+{a}}}+{x}+{a}={x}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} } \\ $$$$\frac{\mathrm{1}}{\:\sqrt{{x}+{a}}}+\left(\sqrt{{x}+{a}}\right)^{\mathrm{2}} ={x}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} } \\ $$$$\left(\sqrt{{x}+{a}}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{{x}^{\mathrm{2}} }={x}−\frac{\mathrm{1}}{\:\sqrt{{x}+{a}}} \\ $$$$\left[\left({x}\sqrt{{x}+{a}}\right)^{\mathrm{2}} −\mathrm{1}\right]\sqrt{{x}+{a}}={x}^{\mathrm{2}} \left({x}\sqrt{{x}+{a}}−\mathrm{1}\right) \\ $$$$\left({x}\sqrt{{x}+{a}}−\mathrm{1}\right)\left({x}\sqrt{{x}+{a}}+\mathrm{1}\right)\sqrt{{x}+{a}}={x}^{\mathrm{2}} \left({x}\sqrt{{x}+{a}}−\mathrm{1}\right) \\ $$$$\left({x}\sqrt{{x}+{a}}−\mathrm{1}\right)\left({ax}+\sqrt{{x}+{a}}\right)=\mathrm{0} \\ $$$$ \\ $$$$\underline{{case}\:\mathrm{1}:\:{x}\sqrt{{x}+{a}}−\mathrm{1}=\mathrm{0}} \\ $$$$\sqrt{{x}+{a}}=\frac{\mathrm{1}}{{x}}\:\Rightarrow{x}>\mathrm{0} \\ $$$$\frac{\mathrm{1}}{{x}^{\mathrm{2}} }={x}+{a} \\ $$$$\frac{\mathrm{1}}{{x}^{\mathrm{3}} \:}−\frac{{a}}{{x}}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\frac{\mathrm{1}}{{x}}=\mathrm{2}\sqrt{\frac{{a}}{\mathrm{3}}}\:\mathrm{sin}\:\left(−\frac{\mathrm{1}}{\mathrm{3}}\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}{a}\sqrt{{a}}}\:+\frac{\mathrm{2}{k}\pi}{\mathrm{3}}\right)\:{with}\:{k}=\mathrm{0},\mathrm{1},\mathrm{2} \\ $$$${since}\:{x}>\mathrm{0}, \\ $$$$\Rightarrow{x}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}\sqrt{\sqrt{\mathrm{5}}+\mathrm{1}}\:\mathrm{sin}\:\left(\frac{\mathrm{2}\pi}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{3}}\:\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}\left(\sqrt{\mathrm{5}}+\mathrm{1}\right)\sqrt{\sqrt{\mathrm{5}}+\mathrm{1}}}\right)}\approx\mathrm{0}.\mathrm{516239} \\ $$$$ \\ $$$$\underline{{case}\:\mathrm{2}:\:{ax}+\sqrt{{x}+{a}}=\mathrm{0}} \\ $$$$\sqrt{{x}+{a}}=−{ax}\:\Rightarrow{x}<\mathrm{0} \\ $$$${a}^{\mathrm{2}} {x}^{\mathrm{2}} −{x}−{a}=\mathrm{0} \\ $$$$\Rightarrow{x}=\frac{\mathrm{1}−\sqrt{\mathrm{1}+\mathrm{4}{a}^{\mathrm{3}} }}{\mathrm{2}{a}^{\mathrm{2}} }=\frac{\mathrm{1}−\sqrt{\mathrm{65}+\mathrm{32}\sqrt{\mathrm{5}}}}{\mathrm{4}\left(\mathrm{3}+\sqrt{\mathrm{5}}\right)}\approx−\mathrm{0}.\mathrm{510194} \\ $$
Commented by Tawa11 last updated on 19/Dec/21
$$\mathrm{Great}\:\mathrm{sir} \\ $$