Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 131363 by shaker last updated on 04/Feb/21

Commented by MJS_new last updated on 04/Feb/21

I don′t think we can exactly solve this. we  need the roots of x^3 +3x+1 and in the next  step the roots of a polynome of degree 4 if  we are smart... otherwise degree 6

$$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{think}\:\mathrm{we}\:\mathrm{can}\:\mathrm{exactly}\:\mathrm{solve}\:\mathrm{this}.\:\mathrm{we} \\ $$$$\mathrm{need}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:{x}^{\mathrm{3}} +\mathrm{3}{x}+\mathrm{1}\:\mathrm{and}\:\mathrm{in}\:\mathrm{the}\:\mathrm{next} \\ $$$$\mathrm{step}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{a}\:\mathrm{polynome}\:\mathrm{of}\:\mathrm{degree}\:\mathrm{4}\:\mathrm{if} \\ $$$$\mathrm{we}\:\mathrm{are}\:\mathrm{smart}...\:\mathrm{otherwise}\:\mathrm{degree}\:\mathrm{6} \\ $$

Commented by MJS_new last updated on 04/Feb/21

try these to see the path  (1) x^3 +3x+1=(x−a)(x−b)(x−c)       [in our case a∈R; b, c ∉R]  (2) x^3 +3x+1=(x−a)(x^2 +ax−(1/a))       [a∈R]  in both cases you can use  t=x+(√(x^2 +1)) ⇔ x=((t^2 −1)/(2t)) → dx=((√(x^2 +1))/t)dt  it′s possible to solve it but exactly? no.

$$\mathrm{try}\:\mathrm{these}\:\mathrm{to}\:\mathrm{see}\:\mathrm{the}\:\mathrm{path} \\ $$$$\left(\mathrm{1}\right)\:{x}^{\mathrm{3}} +\mathrm{3}{x}+\mathrm{1}=\left({x}−{a}\right)\left({x}−{b}\right)\left({x}−{c}\right) \\ $$$$\:\:\:\:\:\left[\mathrm{in}\:\mathrm{our}\:\mathrm{case}\:{a}\in\mathbb{R};\:{b},\:{c}\:\notin\mathbb{R}\right] \\ $$$$\left(\mathrm{2}\right)\:{x}^{\mathrm{3}} +\mathrm{3}{x}+\mathrm{1}=\left({x}−{a}\right)\left({x}^{\mathrm{2}} +{ax}−\frac{\mathrm{1}}{{a}}\right) \\ $$$$\:\:\:\:\:\left[{a}\in\mathbb{R}\right] \\ $$$$\mathrm{in}\:\mathrm{both}\:\mathrm{cases}\:\mathrm{you}\:\mathrm{can}\:\mathrm{use} \\ $$$${t}={x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\:\Leftrightarrow\:{x}=\frac{{t}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}{t}}\:\rightarrow\:{dx}=\frac{\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}{{t}}{dt} \\ $$$$\mathrm{it}'\mathrm{s}\:\mathrm{possible}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{it}\:\mathrm{but}\:\mathrm{exactly}?\:\mathrm{no}. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com