Question Number 78108 by mr W last updated on 14/Jan/20
Commented by mr W last updated on 14/Jan/20
$${find}\:{the}\:{relation}\:{between}\:{a}\:{and}\:{c}. \\ $$$$ \\ $$$$\left({i}\:{reposted}\:{the}\:{question}\:{here}\:{to}\:{avoid}\right. \\ $$$${that}\:{the}\:{solution}\:{will}\:{be}\:{deleted}\:{by} \\ $$$$\left.{someone}.\right) \\ $$
Answered by mr W last updated on 14/Jan/20
![(i)+(ii): 32((c/a)+(a/c))=((x/a))^5 +5((x/a))^3 +10((x/a))+10((a/x))+5((a/x))^3 +((a/x))^5 32((c/a)+(a/c))=((x/a)+(a/x))^5 ⇒(x/a)+(a/x)=2((c/a)+(a/c))^(1/5) ...(I) (i)−(ii): 32((c/a)−(a/c))=((x/a))^5 −5((x/a))^3 +10((x/a))−10((a/x))+5((a/x))^3 −((a/x))^5 32((c/a)−(a/c))=((x/a)−(a/x))^5 ⇒(x/a)−(a/x)=2((c/a)−(a/c))^(1/5) ...(II) (I)^2 −(II)^2 : 4=4((c/a)+(a/c))^(2/5) −4((c/a)−(a/c))^(2/5) ⇒((c/a)+(a/c))^(2/5) −((c/a)−(a/c))^(2/5) =1 or ⇒((((c/a))^2 +((a/c))^2 +2))^(1/5) −((((c/a))^2 +((a/c))^2 −2))^(1/5) =1 let ((c/a))^2 +((a/c))^2 −2=s ⇒((s+4))^(1/5) −(s)^(1/5) =1 ⇒((s+4))^(1/5) =(s)^(1/5) +1 ⇒s+4=s+5((s)^(1/5) )^4 +10((s)^(1/5) )^3 +10((s)^(1/5) )^2 +5((s)^(1/5) )+1 ⇒⇒5((s)^(1/5) )^4 +10((s)^(1/5) )^3 +10((s)^(1/5) )^2 +5((s)^(1/5) )−3=0 let t=(s)^(1/5) =(( ((c/a))^2 +((a/c))^2 −2))^(1/5) ⇒5t^4 +10t^3 +10t^2 +5t−3=0 ⇒5(t^2 +t)^2 +5(t^2 +t)−3=0 ⇒t^2 (t+1)^2 +t(t+1)−(3/5)=0 ⇒t^2 (t+1)^2 +t(t+1)+(1/4)−((3/5)+(1/4))=0 ⇒[t(t+1)+(1/2)]^2 =(((√(85))/(10)))^2 ⇒t(t+1)+(1/2)=((√(85))/(10)) ⇒t^2 +t+(1/2)−((√(85))/(10))=0 ⇒t=(1/4)(−1±(√(((8(√(85)))/5)−7))) ⇒(( ((c/a))^2 +((a/c))^2 −2))^(1/5) =(1/4)(−1+(√(((8(√(85)))/5)−7))) ⇒ (c/a)−(a/c)=±(1/2^5 )(−1+(√(((8(√(85)))/5)−7)))^(5/2) =μ ((c/a))^2 −μ((c/a))−1=0 ⇒(c/a)=(1/2)(μ±(√(μ^2 +4))) ⇒(c/a)=±(1/(64)){(−1+(√(((8(√(85)))/5)−7)))^(5/2) ±(√((−1+(√(((8(√(85)))/5)−7)))^5 +4096))} or (c/a)≈±0.971516, ±1.029319](https://www.tinkutara.com/question/Q78114.png)
$$\left({i}\right)+\left({ii}\right): \\ $$$$\mathrm{32}\left(\frac{{c}}{{a}}+\frac{{a}}{{c}}\right)=\left(\frac{{x}}{{a}}\right)^{\mathrm{5}} +\mathrm{5}\left(\frac{{x}}{{a}}\right)^{\mathrm{3}} +\mathrm{10}\left(\frac{{x}}{{a}}\right)+\mathrm{10}\left(\frac{{a}}{{x}}\right)+\mathrm{5}\left(\frac{{a}}{{x}}\right)^{\mathrm{3}} +\left(\frac{{a}}{{x}}\right)^{\mathrm{5}} \\ $$$$\mathrm{32}\left(\frac{{c}}{{a}}+\frac{{a}}{{c}}\right)=\left(\frac{{x}}{{a}}+\frac{{a}}{{x}}\right)^{\mathrm{5}} \\ $$$$\Rightarrow\frac{{x}}{{a}}+\frac{{a}}{{x}}=\mathrm{2}\left(\frac{{c}}{{a}}+\frac{{a}}{{c}}\right)^{\frac{\mathrm{1}}{\mathrm{5}}} \:\:\:…\left({I}\right) \\ $$$$\left({i}\right)−\left({ii}\right): \\ $$$$\mathrm{32}\left(\frac{{c}}{{a}}−\frac{{a}}{{c}}\right)=\left(\frac{{x}}{{a}}\right)^{\mathrm{5}} −\mathrm{5}\left(\frac{{x}}{{a}}\right)^{\mathrm{3}} +\mathrm{10}\left(\frac{{x}}{{a}}\right)−\mathrm{10}\left(\frac{{a}}{{x}}\right)+\mathrm{5}\left(\frac{{a}}{{x}}\right)^{\mathrm{3}} −\left(\frac{{a}}{{x}}\right)^{\mathrm{5}} \\ $$$$\mathrm{32}\left(\frac{{c}}{{a}}−\frac{{a}}{{c}}\right)=\left(\frac{{x}}{{a}}−\frac{{a}}{{x}}\right)^{\mathrm{5}} \\ $$$$\Rightarrow\frac{{x}}{{a}}−\frac{{a}}{{x}}=\mathrm{2}\left(\frac{{c}}{{a}}−\frac{{a}}{{c}}\right)^{\frac{\mathrm{1}}{\mathrm{5}}} \:\:\:\:\:…\left({II}\right) \\ $$$$ \\ $$$$\left({I}\right)^{\mathrm{2}} −\left({II}\right)^{\mathrm{2}} : \\ $$$$\mathrm{4}=\mathrm{4}\left(\frac{{c}}{{a}}+\frac{{a}}{{c}}\right)^{\frac{\mathrm{2}}{\mathrm{5}}} −\mathrm{4}\left(\frac{{c}}{{a}}−\frac{{a}}{{c}}\right)^{\frac{\mathrm{2}}{\mathrm{5}}} \\ $$$$\Rightarrow\left(\frac{{c}}{{a}}+\frac{{a}}{{c}}\right)^{\frac{\mathrm{2}}{\mathrm{5}}} −\left(\frac{{c}}{{a}}−\frac{{a}}{{c}}\right)^{\frac{\mathrm{2}}{\mathrm{5}}} =\mathrm{1} \\ $$$${or} \\ $$$$\Rightarrow\sqrt[{\mathrm{5}}]{\left(\frac{{c}}{{a}}\right)^{\mathrm{2}} +\left(\frac{{a}}{{c}}\right)^{\mathrm{2}} +\mathrm{2}}−\sqrt[{\mathrm{5}}]{\left(\frac{{c}}{{a}}\right)^{\mathrm{2}} +\left(\frac{{a}}{{c}}\right)^{\mathrm{2}} −\mathrm{2}}=\mathrm{1} \\ $$$${let}\:\left(\frac{{c}}{{a}}\right)^{\mathrm{2}} +\left(\frac{{a}}{{c}}\right)^{\mathrm{2}} −\mathrm{2}={s} \\ $$$$\Rightarrow\sqrt[{\mathrm{5}}]{{s}+\mathrm{4}}−\sqrt[{\mathrm{5}}]{{s}}=\mathrm{1} \\ $$$$\Rightarrow\sqrt[{\mathrm{5}}]{{s}+\mathrm{4}}=\sqrt[{\mathrm{5}}]{{s}}+\mathrm{1} \\ $$$$\Rightarrow{s}+\mathrm{4}={s}+\mathrm{5}\left(\sqrt[{\mathrm{5}}]{{s}}\right)^{\mathrm{4}} +\mathrm{10}\left(\sqrt[{\mathrm{5}}]{{s}}\right)^{\mathrm{3}} +\mathrm{10}\left(\sqrt[{\mathrm{5}}]{{s}}\right)^{\mathrm{2}} +\mathrm{5}\left(\sqrt[{\mathrm{5}}]{{s}}\right)+\mathrm{1} \\ $$$$\Rightarrow\Rightarrow\mathrm{5}\left(\sqrt[{\mathrm{5}}]{{s}}\right)^{\mathrm{4}} +\mathrm{10}\left(\sqrt[{\mathrm{5}}]{{s}}\right)^{\mathrm{3}} +\mathrm{10}\left(\sqrt[{\mathrm{5}}]{{s}}\right)^{\mathrm{2}} +\mathrm{5}\left(\sqrt[{\mathrm{5}}]{{s}}\right)−\mathrm{3}=\mathrm{0} \\ $$$${let}\:{t}=\sqrt[{\mathrm{5}}]{{s}}=\sqrt[{\mathrm{5}}]{\:\left(\frac{{c}}{{a}}\right)^{\mathrm{2}} +\left(\frac{{a}}{{c}}\right)^{\mathrm{2}} −\mathrm{2}} \\ $$$$\Rightarrow\mathrm{5}{t}^{\mathrm{4}} +\mathrm{10}{t}^{\mathrm{3}} +\mathrm{10}{t}^{\mathrm{2}} +\mathrm{5}{t}−\mathrm{3}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{5}\left({t}^{\mathrm{2}} +{t}\right)^{\mathrm{2}} +\mathrm{5}\left({t}^{\mathrm{2}} +{t}\right)−\mathrm{3}=\mathrm{0} \\ $$$$\Rightarrow{t}^{\mathrm{2}} \left({t}+\mathrm{1}\right)^{\mathrm{2}} +{t}\left({t}+\mathrm{1}\right)−\frac{\mathrm{3}}{\mathrm{5}}=\mathrm{0} \\ $$$$\Rightarrow{t}^{\mathrm{2}} \left({t}+\mathrm{1}\right)^{\mathrm{2}} +{t}\left({t}+\mathrm{1}\right)+\frac{\mathrm{1}}{\mathrm{4}}−\left(\frac{\mathrm{3}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{4}}\right)=\mathrm{0} \\ $$$$\Rightarrow\left[{t}\left({t}+\mathrm{1}\right)+\frac{\mathrm{1}}{\mathrm{2}}\right]^{\mathrm{2}} =\left(\frac{\sqrt{\mathrm{85}}}{\mathrm{10}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow{t}\left({t}+\mathrm{1}\right)+\frac{\mathrm{1}}{\mathrm{2}}=\frac{\sqrt{\mathrm{85}}}{\mathrm{10}} \\ $$$$\Rightarrow{t}^{\mathrm{2}} +{t}+\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{85}}}{\mathrm{10}}=\mathrm{0} \\ $$$$\Rightarrow{t}=\frac{\mathrm{1}}{\mathrm{4}}\left(−\mathrm{1}\pm\sqrt{\frac{\mathrm{8}\sqrt{\mathrm{85}}}{\mathrm{5}}−\mathrm{7}}\right) \\ $$$$\Rightarrow\sqrt[{\mathrm{5}}]{\:\left(\frac{{c}}{{a}}\right)^{\mathrm{2}} +\left(\frac{{a}}{{c}}\right)^{\mathrm{2}} −\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{4}}\left(−\mathrm{1}+\sqrt{\frac{\mathrm{8}\sqrt{\mathrm{85}}}{\mathrm{5}}−\mathrm{7}}\right) \\ $$$$\Rightarrow\:\frac{{c}}{{a}}−\frac{{a}}{{c}}=\pm\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{5}} }\left(−\mathrm{1}+\sqrt{\frac{\mathrm{8}\sqrt{\mathrm{85}}}{\mathrm{5}}−\mathrm{7}}\right)^{\mathrm{5}/\mathrm{2}} =\mu \\ $$$$\left(\frac{{c}}{{a}}\right)^{\mathrm{2}} −\mu\left(\frac{{c}}{{a}}\right)−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\frac{{c}}{{a}}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mu\pm\sqrt{\mu^{\mathrm{2}} +\mathrm{4}}\right) \\ $$$$\Rightarrow\frac{{c}}{{a}}=\pm\frac{\mathrm{1}}{\mathrm{64}}\left\{\left(−\mathrm{1}+\sqrt{\frac{\mathrm{8}\sqrt{\mathrm{85}}}{\mathrm{5}}−\mathrm{7}}\right)^{\mathrm{5}/\mathrm{2}} \pm\sqrt{\left(−\mathrm{1}+\sqrt{\frac{\mathrm{8}\sqrt{\mathrm{85}}}{\mathrm{5}}−\mathrm{7}}\right)^{\mathrm{5}} +\mathrm{4096}}\right\} \\ $$$${or} \\ $$$$\frac{{c}}{{a}}\approx\pm\mathrm{0}.\mathrm{971516},\:\pm\mathrm{1}.\mathrm{029319} \\ $$
Commented by TawaTawa last updated on 14/Jan/20
$$\mathrm{Weldone}\:\mathrm{sir}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}. \\ $$