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Question Number 197980 by mr W last updated on 07/Oct/23

Commented by mr W last updated on 07/Oct/23

please exact solution!

$${please}\:{exact}\:{solution}! \\ $$

Commented by Blackpanther last updated on 07/Oct/23

0.196....only real solution

Commented by Frix last updated on 07/Oct/23

After trying for a while: x=2^(3/5) −2^(2/5) Not sure how to get it analytically

$$\mathrm{After}\:\mathrm{trying}\:\mathrm{for}\:\mathrm{a}\:\mathrm{while}: \\ $$$${x}=\mathrm{2}^{\frac{\mathrm{3}}{\mathrm{5}}} −\mathrm{2}^{\frac{\mathrm{2}}{\mathrm{5}}} \\ $$$$\mathrm{Not}\:\mathrm{sure}\:\mathrm{how}\:\mathrm{to}\:\mathrm{get}\:\mathrm{it}\:\mathrm{analytically} \\ $$

Commented by mr W last updated on 07/Oct/23

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Answered by mr W last updated on 07/Oct/23

for this kind of quintic equations we can try with a substitution like x=t+(a/t), here with x=t−(2/t). then we get t^5 −((32)/t^5 )−4=0, or t^(10) −4t^5 −32=0 (t^5 −8)(t^5 +4)=0 ⇒t^5 =8 ⇒t=(8)^(1/5) ⇒x=(8)^(1/5) −(2/( (8)^(1/5) ))=2^(3/5) −2^(2/5) ≈0.196209

$${for}\:{this}\:{kind}\:{of}\:{quintic}\:{equations} \\ $$$${we}\:{can}\:{try}\:{with}\:{a}\:{substitution}\:{like} \\ $$$${x}={t}+\frac{{a}}{{t}},\:{here}\:{with}\:{x}={t}−\frac{\mathrm{2}}{{t}}.\:{then} \\ $$$${we}\:{get} \\ $$$${t}^{\mathrm{5}} −\frac{\mathrm{32}}{{t}^{\mathrm{5}} }−\mathrm{4}=\mathrm{0},\:{or} \\ $$$${t}^{\mathrm{10}} −\mathrm{4}{t}^{\mathrm{5}} −\mathrm{32}=\mathrm{0} \\ $$$$\left({t}^{\mathrm{5}} −\mathrm{8}\right)\left({t}^{\mathrm{5}} +\mathrm{4}\right)=\mathrm{0} \\ $$$$\Rightarrow{t}^{\mathrm{5}} =\mathrm{8}\:\Rightarrow{t}=\sqrt[{\mathrm{5}}]{\mathrm{8}} \\ $$$$\Rightarrow{x}=\sqrt[{\mathrm{5}}]{\mathrm{8}}−\frac{\mathrm{2}}{\:\sqrt[{\mathrm{5}}]{\mathrm{8}}}=\mathrm{2}^{\frac{\mathrm{3}}{\mathrm{5}}} −\mathrm{2}^{\frac{\mathrm{2}}{\mathrm{5}}} \approx\mathrm{0}.\mathrm{196209} \\ $$

Commented by mr W last updated on 07/Oct/23

it can be easily checked that the eqn. has one and only one real root between 0 and 1. f(0)=−4<0 f(1)=27>0 f′(x)=5x^4 +30x^2 +20>0

$${it}\:{can}\:{be}\:{easily}\:{checked}\:{that}\:{the}\:{eqn}. \\ $$$${has}\:{one}\:{and}\:{only}\:{one}\:{real}\:{root}\: \\ $$$${between}\:\mathrm{0}\:{and}\:\mathrm{1}. \\ $$$${f}\left(\mathrm{0}\right)=−\mathrm{4}<\mathrm{0} \\ $$$${f}\left(\mathrm{1}\right)=\mathrm{27}>\mathrm{0} \\ $$$${f}'\left({x}\right)=\mathrm{5}{x}^{\mathrm{4}} +\mathrm{30}{x}^{\mathrm{2}} +\mathrm{20}>\mathrm{0} \\ $$

Commented by mr W last updated on 07/Oct/23

yes. only solvable if p^2 =5q.

$${yes}.\:{only}\:{solvable}\:{if}\:{p}^{\mathrm{2}} =\mathrm{5}{q}. \\ $$

Commented by Frix last updated on 07/Oct/23

But this is only good luck. x^5 +px^3 +qx+r=0 x=t+(a/t) ⇒ t^5 +(5a+p)t^3 +(10a^2 +3ap+q)t+((a(10a^2 +3ap+q))/t)+((a^3 (5a+p))/t^3 )+(a^5 /t^5 )+r=0 It works only if q=(p^2 /5) ⇒ a=−(p/5)=±((√(5q))/5) ⇒ t^5 +(a^5 /t^5 )+r=0

$$\mathrm{But}\:\mathrm{this}\:\mathrm{is}\:\mathrm{only}\:\mathrm{good}\:\mathrm{luck}. \\ $$$${x}^{\mathrm{5}} +{px}^{\mathrm{3}} +{qx}+{r}=\mathrm{0} \\ $$$${x}={t}+\frac{{a}}{{t}}\:\Rightarrow \\ $$$${t}^{\mathrm{5}} +\left(\mathrm{5}{a}+{p}\right){t}^{\mathrm{3}} +\left(\mathrm{10}{a}^{\mathrm{2}} +\mathrm{3}{ap}+{q}\right){t}+\frac{{a}\left(\mathrm{10}{a}^{\mathrm{2}} +\mathrm{3}{ap}+{q}\right)}{{t}}+\frac{{a}^{\mathrm{3}} \left(\mathrm{5}{a}+{p}\right)}{{t}^{\mathrm{3}} }+\frac{{a}^{\mathrm{5}} }{{t}^{\mathrm{5}} }+{r}=\mathrm{0} \\ $$$$\mathrm{It}\:\mathrm{works}\:\mathrm{only}\:\mathrm{if} \\ $$$${q}=\frac{{p}^{\mathrm{2}} }{\mathrm{5}}\:\Rightarrow\:{a}=−\frac{{p}}{\mathrm{5}}=\pm\frac{\sqrt{\mathrm{5}{q}}}{\mathrm{5}} \\ $$$$\Rightarrow \\ $$$${t}^{\mathrm{5}} +\frac{{a}^{\mathrm{5}} }{{t}^{\mathrm{5}} }+{r}=\mathrm{0} \\ $$

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