Question Number 201139 by cherokeesay last updated on 30/Nov/23
Commented by Frix last updated on 30/Nov/23
$${x}\approx\mathrm{6395}.\mathrm{12283}\wedge{y}\approx\mathrm{171}.\mathrm{458282} \\ $$$$\mathrm{Exact}\:\mathrm{solution}: \\ $$$${x}=\mathrm{32}\left(\mathrm{8}\left(\mathrm{5}{r}^{\mathrm{2}} +\mathrm{6}{r}+\mathrm{9}\right)\sqrt{{r}−\mathrm{1}}+\mathrm{16}{r}^{\mathrm{2}} +\mathrm{29}{r}+\mathrm{38}\right) \\ $$$${y}=\mathrm{16}\left(\mathrm{4}\left({r}^{\mathrm{2}} +\mathrm{4}{r}−\mathrm{4}\right)\sqrt{{r}−\mathrm{1}}+\mathrm{16}{r}^{\mathrm{2}} −\mathrm{4}{r}−\mathrm{15}\right) \\ $$$${r}=\sqrt[{\mathrm{3}}]{\mathrm{2}} \\ $$$$\mathrm{I}\:\mathrm{got}\:\mathrm{this}\:\mathrm{by}\:\mathrm{substituting} \\ $$$${x}={p}^{\mathrm{4}} \wedge{y}={p}^{\mathrm{4}} {q}^{\mathrm{4}} \wedge{p},\:{q}\:>\mathrm{0} \\ $$$$\mathrm{but}\:\mathrm{it}'\mathrm{s}\:\mathrm{too}\:\mathrm{much}\:\mathrm{work}\:\mathrm{to}\:\mathrm{type}\:\mathrm{it}\:\mathrm{here}. \\ $$
Answered by ajfour last updated on 01/Dec/23
![(√x)=p , (√y)=q , a=(1/4) , z=((2p+q)/(p^2 +q^2 )) p(a−((2p+q)/(p^2 +q^2 )))^2 =4 q(a+((2p+q)/(p^2 +q^2 )))^2 =1 a−z=(2/( (√p))) a+z=(1/( (√q))) 2z=(((√p)−2(√q))/( (√p)(√q)))=2(((2p+q)/( p^2 +q^2 ))) 2a=(((√p)+2(√q))/( (√p)(√q))) a^2 −z^2 =(2/( (√p)(√q))) z{((16)/((a−z)^4 ))+(1/((a+z)^4 ))}=(8/((a−z)^2 ))+(1/((a+z)^2 )) 16z(a+z)^4 +z(a−z)^4 =8(a−z)^2 (a+z)^4 +(a+z)^2 (a−z)^4 or if z=acos 2θ =a(h−k) h+k=1 4(h^2 −k^2 ){(1/k^4 )+(1/(16h^4 ))} =(2/k^2 )+(1/(4h^2 )) ⇒ ((4h^2 )/k^4 )−(k^2 /(4h^4 ))+(1/(4h^2 ))−(4/k^2 )=(2/k^2 )+(1/(4h^2 )) ⇒ ((4h^2 )/k^4 )−(k^2 /(4h^4 ))=(6/k^2 ) 16h^6 −k^6 =24k^2 h^4 say tan θ=(k/h)=t t^6 +24t^2 −16=0 ⇒ t^2 =2((4)^(1/3) −(2)^(1/3) ) z=acos 2θ (√x)=p=(4/((a−z)^2 ))=(1/(a^2 sin^4 θ))=((16)/k^2 ) (√y)=q=(1/((a+z)^2 ))=(1/(4a^2 cos^4 θ))=(4/h^2 ) (k/h)=t ; h+k=1 ⇒ h=(1/(1+t))=(1/(1+2((4)^(1/3) −(2)^(1/3) ))) & k=((2((4)^(1/3) −(2)^(1/3) ))/(1+2((4)^(1/3) −(2)^(1/3) ))) (√x)=16[((1+2((4)^(1/3) −(2)^(1/3) ))/(2((4)^(1/3) −(2)^(1/3) )))]^2 ⇒ x≈ 69121.4399 (√y)=4[1+2((4)^(1/3) −(2)^(1/3) )]^2 ≈ 120.024524](https://www.tinkutara.com/question/Q201148.png)
$$\sqrt{{x}}={p}\:\:\:\:,\:\:\sqrt{{y}}={q}\:\:,\:\:{a}=\frac{\mathrm{1}}{\mathrm{4}}\:,\:{z}=\frac{\mathrm{2}{p}+{q}}{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} } \\ $$$${p}\left({a}−\frac{\mathrm{2}{p}+{q}}{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} }\right)^{\mathrm{2}} =\mathrm{4} \\ $$$${q}\left({a}+\frac{\mathrm{2}{p}+{q}}{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} }\right)^{\mathrm{2}} =\mathrm{1} \\ $$$${a}−{z}=\frac{\mathrm{2}}{\:\sqrt{{p}}} \\ $$$${a}+{z}=\frac{\mathrm{1}}{\:\sqrt{{q}}} \\ $$$$\mathrm{2}{z}=\frac{\sqrt{{p}}−\mathrm{2}\sqrt{{q}}}{\:\sqrt{{p}}\sqrt{{q}}}=\mathrm{2}\left(\frac{\mathrm{2}{p}+{q}}{\:{p}^{\mathrm{2}} +{q}^{\mathrm{2}} }\right) \\ $$$$\mathrm{2}{a}=\frac{\sqrt{{p}}+\mathrm{2}\sqrt{{q}}}{\:\sqrt{{p}}\sqrt{{q}}} \\ $$$${a}^{\mathrm{2}} −{z}^{\mathrm{2}} =\frac{\mathrm{2}}{\:\sqrt{{p}}\sqrt{{q}}} \\ $$$${z}\left\{\frac{\mathrm{16}}{\left({a}−{z}\right)^{\mathrm{4}} }+\frac{\mathrm{1}}{\left({a}+{z}\right)^{\mathrm{4}} }\right\}=\frac{\mathrm{8}}{\left({a}−{z}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\left({a}+{z}\right)^{\mathrm{2}} } \\ $$$$\mathrm{16}{z}\left({a}+{z}\right)^{\mathrm{4}} +{z}\left({a}−{z}\right)^{\mathrm{4}} \\ $$$$\:\:=\mathrm{8}\left({a}−{z}\right)^{\mathrm{2}} \left({a}+{z}\right)^{\mathrm{4}} +\left({a}+{z}\right)^{\mathrm{2}} \left({a}−{z}\right)^{\mathrm{4}} \\ $$$${or}\:\:{if}\:{z}={a}\mathrm{cos}\:\mathrm{2}\theta\:\:={a}\left({h}−{k}\right) \\ $$$${h}+{k}=\mathrm{1} \\ $$$$\mathrm{4}\left({h}^{\mathrm{2}} −{k}^{\mathrm{2}} \right)\left\{\frac{\mathrm{1}}{{k}^{\mathrm{4}} }+\frac{\mathrm{1}}{\mathrm{16}{h}^{\mathrm{4}} }\right\} \\ $$$$\:\:\:=\frac{\mathrm{2}}{{k}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{4}{h}^{\mathrm{2}} } \\ $$$$\Rightarrow\:\:\frac{\mathrm{4}{h}^{\mathrm{2}} }{{k}^{\mathrm{4}} }−\frac{{k}^{\mathrm{2}} }{\mathrm{4}{h}^{\mathrm{4}} }+\frac{\mathrm{1}}{\mathrm{4}{h}^{\mathrm{2}} }−\frac{\mathrm{4}}{{k}^{\mathrm{2}} }=\frac{\mathrm{2}}{{k}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{4}{h}^{\mathrm{2}} } \\ $$$$\Rightarrow\:\:\frac{\mathrm{4}{h}^{\mathrm{2}} }{{k}^{\mathrm{4}} }−\frac{{k}^{\mathrm{2}} }{\mathrm{4}{h}^{\mathrm{4}} }=\frac{\mathrm{6}}{{k}^{\mathrm{2}} } \\ $$$$\mathrm{16}{h}^{\mathrm{6}} −{k}^{\mathrm{6}} =\mathrm{24}{k}^{\mathrm{2}} {h}^{\mathrm{4}} \\ $$$${say}\:\:\mathrm{tan}\:\theta=\frac{{k}}{{h}}={t} \\ $$$${t}^{\mathrm{6}} +\mathrm{24}{t}^{\mathrm{2}} −\mathrm{16}=\mathrm{0} \\ $$$$\Rightarrow\:\:{t}^{\mathrm{2}} =\mathrm{2}\left(\sqrt[{\mathrm{3}}]{\mathrm{4}}−\sqrt[{\mathrm{3}}]{\mathrm{2}}\right) \\ $$$${z}={a}\mathrm{cos}\:\mathrm{2}\theta \\ $$$$\sqrt{{x}}={p}=\frac{\mathrm{4}}{\left({a}−{z}\right)^{\mathrm{2}} }=\frac{\mathrm{1}}{{a}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{4}} \theta}=\frac{\mathrm{16}}{{k}^{\mathrm{2}} } \\ $$$$\sqrt{{y}}={q}=\frac{\mathrm{1}}{\left({a}+{z}\right)^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{4}{a}^{\mathrm{2}} \mathrm{cos}\:^{\mathrm{4}} \theta}=\frac{\mathrm{4}}{{h}^{\mathrm{2}} } \\ $$$$\:\:\frac{{k}}{{h}}={t}\:\:\:\:\:\:;\:\:\:{h}+{k}=\mathrm{1} \\ $$$$\Rightarrow\:\:\:{h}=\frac{\mathrm{1}}{\mathrm{1}+{t}}=\frac{\mathrm{1}}{\mathrm{1}+\mathrm{2}\left(\sqrt[{\mathrm{3}}]{\mathrm{4}}−\sqrt[{\mathrm{3}}]{\mathrm{2}}\right)} \\ $$$$\&\:\:\:\:\:{k}=\frac{\mathrm{2}\left(\sqrt[{\mathrm{3}}]{\mathrm{4}}−\sqrt[{\mathrm{3}}]{\mathrm{2}}\right)}{\mathrm{1}+\mathrm{2}\left(\sqrt[{\mathrm{3}}]{\mathrm{4}}−\sqrt[{\mathrm{3}}]{\mathrm{2}}\right)} \\ $$$$\sqrt{{x}}=\mathrm{16}\left[\frac{\mathrm{1}+\mathrm{2}\left(\sqrt[{\mathrm{3}}]{\mathrm{4}}−\sqrt[{\mathrm{3}}]{\mathrm{2}}\right)}{\mathrm{2}\left(\sqrt[{\mathrm{3}}]{\mathrm{4}}−\sqrt[{\mathrm{3}}]{\mathrm{2}}\right)}\right]^{\mathrm{2}} \:\: \\ $$$$\Rightarrow\:\:{x}\approx\:\mathrm{69121}.\mathrm{4399} \\ $$$$\sqrt{{y}}=\mathrm{4}\left[\mathrm{1}+\mathrm{2}\left(\sqrt[{\mathrm{3}}]{\mathrm{4}}−\sqrt[{\mathrm{3}}]{\mathrm{2}}\right)\right]^{\mathrm{2}} \\ $$$$\:\:\:\:\:\approx\:\mathrm{120}.\mathrm{024524} \\ $$
Commented by ajfour last updated on 30/Nov/23
$${but}\:{i}\:{checked}\:\:{my}\:{answers}\:{are} \\ $$$${wrong}! \\ $$