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 141076 by ajfour last updated on 15/May/21

$$t+x=\frac{c}{2}$$$$t^2+x^4=\frac{c^2}{4}$$$$findxort.Given0<c<\frac{2}{3\sqrt{3}}$$

Answered by Rasheed.Sindhi last updated on 15/May/21

$$t-\frac{c}{2}=-x$$$$t^2-\frac{c^2}{4}=-x^4\Rightarrow (t-\frac{c}{2})(t+\frac{c}{2})=-x^4$$$$\Rightarrow (-x)(t+\frac{c}{2})+x^4=0$$$$\Rightarrow x(-t-\frac{c}{2}+x^3)=0$$$$\Rightarrow x=0\mid -t-\frac{c}{2}+x^3=0$$$$x=0\Rightarrow t+0=\frac{c}{2}\Rightarrow t=\frac{c}{2}$$$$(x,t)=(0,\frac{c}{2})$$$$-t-\frac{c}{2}+x^3=0...........\left(i\right)$$$$t-\frac{c}{2}+x=0.............\left(ii\right)$$$$\left(i\right)+\left(ii\right)$$$$x^3+x-c=0\left(★\right)$$$$$$$$\left(i\right)\Rightarrow x^3=t+\frac{c}{2}........\left(iii\right)$$$$\left(ii\right)\Rightarrow x^3={(-t+\frac{c}{2})}^3....\left(iv\right)$$$$t+\frac{c}{2}={(-t+\frac{c}{2})}^3$$$$t+\frac{c}{2}=-t^3+\frac{c^3}{8}+3(-t)\left(\frac{c}{2}\right)(-t+\frac{c}{2})$$$$t+\frac{c}{2}=-t^3+\frac{c^3}{8}+\frac{3}{2}{ct}^2-\frac{3}{2}ct$$$$t^3-\frac{3}{2}{ct}^2+\frac{3}{2}ct-\frac{c^3}{8}+\frac{c}{2}=0\left(★\right)$$$$\left(★\right)\mathcal{D}{on}^{\prime }tknowhowtosolvecubic$$$$equation.$$$$(x,t)=(0,\frac{c}{2})$$$$......$$$$$$

Commented by mr W last updated on 15/May/21

$$x^3+x-c=0\left(★\right)$$$$hasalwaysonerealroot:$$$$x=\sqrt[3]{\sqrt{\frac{1}{27}+\frac{c^2}{4}}+\frac{c}{2}}-\sqrt[3]{\sqrt{\frac{1}{27}+\frac{c^2}{4}}-\frac{c}{4}}$$

Commented by Rasheed.Sindhi last updated on 15/May/21

$$\mathcal{T}h\alpha nksmr\mathcal{W}\mathit{s}\mathit{i}\mathit{r}!$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com