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Question Number 78037 by Pratah last updated on 13/Jan/20

Commented by MJS last updated on 13/Jan/20

see question 77009

$$\mathrm{see}\:\mathrm{question}\:\mathrm{77009} \\ $$

Commented by Pratah last updated on 13/Jan/20

parameter parameters X, where the current value is.

$$\mathrm{parameter}\:\:\mathrm{parameters}\:\mathrm{X},\:\mathrm{where}\:\:\mathrm{the}\:\mathrm{current}\:\mathrm{value}\:\mathrm{is}. \\ $$

Commented by mr W last updated on 14/Jan/20

sir Pratah (BK, Master,etc.): i′d like to give an answer to this question, but i really don′t know what you exactly want. can you give a little explanation what you want?

$${sir}\:{Pratah}\:\left({BK},\:{Master},{etc}.\right): \\ $$$${i}'{d}\:{like}\:{to}\:{give}\:{an}\:{answer}\:{to}\:{this}\:{question}, \\ $$$${but}\:{i}\:{really}\:{don}'{t}\:{know}\:{what}\:{you}\:{exactly} \\ $$$${want}.\:{can}\:{you}\:{give}\:{a}\:{little}\:{explanation} \\ $$$${what}\:{you}\:{want}? \\ $$

Commented by mr W last updated on 14/Jan/20

do you want something like this: (x/a)=((c/a)+(a/c))^(1/5) ±(√([((c/a)+(a/c))^(1/5) +1][((c/a)+(a/c))^(1/5) −1]))

$${do}\:{you}\:{want}\:{something}\:{like}\:{this}: \\ $$$$\frac{{x}}{{a}}=\left(\frac{{c}}{{a}}+\frac{{a}}{{c}}\right)^{\frac{\mathrm{1}}{\mathrm{5}}} \pm\sqrt{\left[\left(\frac{{c}}{{a}}+\frac{{a}}{{c}}\right)^{\frac{\mathrm{1}}{\mathrm{5}}} +\mathrm{1}\right]\left[\left(\frac{{c}}{{a}}+\frac{{a}}{{c}}\right)^{\frac{\mathrm{1}}{\mathrm{5}}} −\mathrm{1}\right]} \\ $$

Commented by Pratah last updated on 14/Jan/20

no sir

$$\mathrm{no}\:\mathrm{sir} \\ $$

Commented by Pratah last updated on 14/Jan/20

?

$$? \\ $$

Commented by mr W last updated on 14/Jan/20

then what is exactly the question?

$${then}\:{what}\:{is}\:{exactly}\:{the}\:{question}? \\ $$

Commented by mr W last updated on 14/Jan/20

sir Pratah (BK, Master,etc.): some minutes ago, as i gave my comment above, i could open the post 77009. after i have given the comment, you have also deleted that old post. it can not be opened anymore. why do you do such things sir?

$${sir}\:{Pratah}\:\left({BK},\:{Master},{etc}.\right): \\ $$$${some}\:{minutes}\:{ago},\:{as}\:{i}\:{gave}\:{my}\:{comment} \\ $$$${above},\:{i}\:{could}\:{open}\:{the}\:{post}\:\mathrm{77009}. \\ $$$${after}\:{i}\:{have}\:{given}\:{the}\:{comment},\:{you} \\ $$$${have}\:{also}\:{deleted}\:{that}\:{old}\:{post}.\:{it}\:{can}\:{not} \\ $$$${be}\:{opened}\:{anymore}. \\ $$$${why}\:\:{do}\:{you}\:{do}\:{such}\:{things}\:{sir}? \\ $$

Commented by Pratah last updated on 14/Jan/20

can you find the relationship between a and c.?

$$\mathrm{can}\:\mathrm{you}\:\mathrm{find}\:\mathrm{the}\:\mathrm{relationship}\:\mathrm{between}\:\mathrm{a}\:\mathrm{and}\:\mathrm{c}.? \\ $$

Commented by MJS last updated on 14/Jan/20

another path let t=(x/a) eliminate c let y=t^4 solve for y by finding the 2 square factors of the polynome

$$\mathrm{another}\:\mathrm{path} \\ $$$$\mathrm{let}\:{t}=\frac{{x}}{{a}} \\ $$$$\mathrm{eliminate}\:{c} \\ $$$$\mathrm{let}\:{y}={t}^{\mathrm{4}} \\ $$$$\mathrm{solve}\:\mathrm{for}\:{y}\:\mathrm{by}\:\mathrm{finding}\:\mathrm{the}\:\mathrm{2}\:\mathrm{square}\:\mathrm{factors}\:\mathrm{of} \\ $$$$\:\:\:\:\:\mathrm{the}\:\mathrm{polynome} \\ $$

Commented by mr W last updated on 14/Jan/20

sir Pratah (BK, Master,etc.): i have posted my answer in a new post. is that answer right? please give substantial feedback.

$${sir}\:{Pratah}\:\left({BK},\:{Master},{etc}.\right): \\ $$$${i}\:{have}\:{posted}\:{my}\:{answer}\:{in}\:{a}\:{new}\: \\ $$$${post}.\:{is}\:{that}\:{answer}\:{right}?\:{please} \\ $$$${give}\:{substantial}\:{feedback}. \\ $$

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