Question-210996

Question Number 210996 by RojaTaniya last updated on 25/Aug/24

Answered by Frix last updated on 26/Aug/24

Only true for the real solution >0: x^3 +2x^2 +5x−1=0 x^3 +6x^2 +9x=4x^2 +4x+1 x(x+3)^2 =(2x+1)^2 x^(1/2) (x+3)=2x+1 x^(3/2) +3x^(1/2) =2x+1 x=1−3x^(1/2) +3x−x^(3/2) x=(1−x^(1/2) )^3 x^(1/3) =1−x^(1/2) x^(1/2) +x^(1/3) =1

$$\mathrm{Only}\:\mathrm{true}\:\mathrm{for}\:\mathrm{the}\:\mathrm{real}\:\mathrm{solution}\:>\mathrm{0}: \\ $$$${x}^{\mathrm{3}} +\mathrm{2}{x}^{\mathrm{2}} +\mathrm{5}{x}−\mathrm{1}=\mathrm{0} \\ $$$${x}^{\mathrm{3}} +\mathrm{6}{x}^{\mathrm{2}} +\mathrm{9}{x}=\mathrm{4}{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{1} \\ $$$${x}\left({x}+\mathrm{3}\right)^{\mathrm{2}} =\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$${x}^{\frac{\mathrm{1}}{\mathrm{2}}} \left({x}+\mathrm{3}\right)=\mathrm{2}{x}+\mathrm{1} \\ $$$${x}^{\frac{\mathrm{3}}{\mathrm{2}}} +\mathrm{3}{x}^{\frac{\mathrm{1}}{\mathrm{2}}} =\mathrm{2}{x}+\mathrm{1} \\ $$$${x}=\mathrm{1}−\mathrm{3}{x}^{\frac{\mathrm{1}}{\mathrm{2}}} +\mathrm{3}{x}−{x}^{\frac{\mathrm{3}}{\mathrm{2}}} \\ $$$${x}=\left(\mathrm{1}−{x}^{\frac{\mathrm{1}}{\mathrm{2}}} \right)^{\mathrm{3}} \\ $$$${x}^{\frac{\mathrm{1}}{\mathrm{3}}} =\mathrm{1}−{x}^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$${x}^{\frac{\mathrm{1}}{\mathrm{2}}} +{x}^{\frac{\mathrm{1}}{\mathrm{3}}} =\mathrm{1} \\ $$

Commented by Rasheed.Sindhi last updated on 26/Aug/24

$$\vee.\:\:\cap\boldsymbol{\mathrm{i}}\subset\in!\:\: \\ $$

Commented by Frix last updated on 26/Aug/24

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Answered by Frix last updated on 26/Aug/24

Another method, not recommended, thus I don′t write it out... (x^3 +2x^2 +5x−1):(x^(1/2) +x^(1/3) −y) (t^(18) +2t^(12) +5t^6 −1):(t^3 +t^2 −y)= =Σ_(j=0) ^(15) (c_j t^j )−((y−1)/(t^3 +t^2 −y)) Σ_(k=0) ^2 (d_k t^k ) ⇒ y=1

$$\mathrm{Another}\:\mathrm{method},\:\mathrm{not}\:\mathrm{recommended},\:\mathrm{thus} \\ $$$$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{write}\:\mathrm{it}\:\mathrm{out}… \\ $$$$\left({x}^{\mathrm{3}} +\mathrm{2}{x}^{\mathrm{2}} +\mathrm{5}{x}−\mathrm{1}\right):\left({x}^{\frac{\mathrm{1}}{\mathrm{2}}} +{x}^{\frac{\mathrm{1}}{\mathrm{3}}} −{y}\right) \\ $$$$\left({t}^{\mathrm{18}} +\mathrm{2}{t}^{\mathrm{12}} +\mathrm{5}{t}^{\mathrm{6}} −\mathrm{1}\right):\left({t}^{\mathrm{3}} +{t}^{\mathrm{2}} −{y}\right)= \\ $$$$=\underset{{j}=\mathrm{0}} {\overset{\mathrm{15}} {\sum}}\:\left({c}_{{j}} {t}^{{j}} \right)−\frac{{y}−\mathrm{1}}{{t}^{\mathrm{3}} +{t}^{\mathrm{2}} −{y}}\:\underset{{k}=\mathrm{0}} {\overset{\mathrm{2}} {\sum}}\:\left({d}_{{k}} {t}^{{k}} \right) \\ $$$$\Rightarrow\:{y}=\mathrm{1} \\ $$

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