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Question Number 153226 by ajfour last updated on 05/Sep/21

Commented by mr W last updated on 05/Sep/21

x^3 +x−c=0 ⇒x=(((√((1/(27))+(c^2 /4)))+(c/2)))^(1/3) −(((√((1/(27))+(c^2 /4)))−(c/2)))^(1/3) any other ways to solve?

$${x}^{\mathrm{3}} +{x}−{c}=\mathrm{0} \\ $$$$\Rightarrow{x}=\sqrt[{\mathrm{3}}]{\sqrt{\frac{\mathrm{1}}{\mathrm{27}}+\frac{{c}^{\mathrm{2}} }{\mathrm{4}}}+\frac{{c}}{\mathrm{2}}}−\sqrt[{\mathrm{3}}]{\sqrt{\frac{\mathrm{1}}{\mathrm{27}}+\frac{{c}^{\mathrm{2}} }{\mathrm{4}}}−\frac{{c}}{\mathrm{2}}} \\ $$$${any}\:{other}\:{ways}\:{to}\:{solve}? \\ $$

Commented by ajfour last updated on 05/Sep/21

Commented by TVTA last updated on 06/Sep/21

x^3 +x−c=0 ⇒x=(((√((1/(27))+(c^2 /4)))+(c/2)))^(1/3) −(((√((1/(27))+(c^2 /4)))−(c/2)))^(1/3) any other ways to solve? x=(((c+(√(c^2 +(4/(27)))))/2))^(1/3) +(1/( (((c+(√(c^2 +(4/(27)))))/2))^(1/3) )) is this correct and ok sir ?

$${x}^{\mathrm{3}} +{x}−{c}=\mathrm{0} \\ $$$$\Rightarrow{x}=\sqrt[{\mathrm{3}}]{\sqrt{\frac{\mathrm{1}}{\mathrm{27}}+\frac{{c}^{\mathrm{2}} }{\mathrm{4}}}+\frac{{c}}{\mathrm{2}}}−\sqrt[{\mathrm{3}}]{\sqrt{\frac{\mathrm{1}}{\mathrm{27}}+\frac{{c}^{\mathrm{2}} }{\mathrm{4}}}−\frac{{c}}{\mathrm{2}}} \\ $$$${any}\:{other}\:{ways}\:{to}\:{solve}? \\ $$$${x}=\sqrt[{\mathrm{3}}]{\frac{{c}+\sqrt{{c}^{\mathrm{2}} +\frac{\mathrm{4}}{\mathrm{27}}}}{\mathrm{2}}}+\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\frac{{c}+\sqrt{{c}^{\mathrm{2}} +\frac{\mathrm{4}}{\mathrm{27}}}}{\mathrm{2}}}} \\ $$$${is}\:\:{this}\:{correct}\:{and}\:{ok}\:{sir}\:? \\ $$

Commented by ajfour last updated on 06/Sep/21

thanks everyone!

$${thanks}\:{everyone}! \\ $$

Answered by talminator2856791 last updated on 06/Sep/21

2x(x^2 −1)+2x = c 2x^3 −2x+2x = c 2x^3 = c x^3 = c∙2^(−1) x = (c∙2^(−1) )^(1/3)

$$\:\mathrm{2}{x}\left({x}^{\mathrm{2}} −\mathrm{1}\right)+\mathrm{2}{x}\:=\:{c} \\ $$$$\:\mathrm{2}{x}^{\mathrm{3}} −\mathrm{2}{x}+\mathrm{2}{x}\:=\:{c} \\ $$$$\:\mathrm{2}{x}^{\mathrm{3}} \:=\:{c} \\ $$$$\:{x}^{\mathrm{3}} \:=\:{c}\centerdot\mathrm{2}^{−\mathrm{1}} \\ $$$$\:{x}\:=\:\left({c}\centerdot\mathrm{2}^{−\mathrm{1}} \right)^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$

Commented by MJS_new last updated on 06/Sep/21

formula for area of trapezoid with sides a∥c and d=b with given height h is (((a+c)h)/2) in this case (((2+2x^2 )x)/2)=c ⇔ x^3 +x=c

$$\mathrm{formula}\:\mathrm{for}\:\mathrm{area}\:\mathrm{of}\:\mathrm{trapezoid}\:\mathrm{with}\:\mathrm{sides} \\ $$$${a}\parallel{c}\:\mathrm{and}\:{d}={b}\:\mathrm{with}\:\mathrm{given}\:\mathrm{height}\:{h}\:\mathrm{is}\:\frac{\left({a}+{c}\right){h}}{\mathrm{2}} \\ $$$$\mathrm{in}\:\mathrm{this}\:\mathrm{case} \\ $$$$\frac{\left(\mathrm{2}+\mathrm{2}{x}^{\mathrm{2}} \right){x}}{\mathrm{2}}={c}\:\Leftrightarrow\:{x}^{\mathrm{3}} +{x}={c} \\ $$

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