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Question Number 218047 by dscm last updated on 27/Mar/25
Determine x: (√(x+3)) + ((x+1))^(1/3) =2
$${Determine}\:{x}: \\ $$$$\sqrt{{x}+\mathrm{3}}\:+\:\sqrt[{\mathrm{3}}]{{x}+\mathrm{1}}\:=\mathrm{2} \\ $$
Answered by Rasheed.Sindhi last updated on 27/Mar/25
(√(x+3)) + ((x+1))^(1/3) =2 ((x+1))^(1/3) =y⇒x=y^3 −1 (√(y^3 +2)) =2−y y^3 +2=y^2 +4−4y y^3 −y^2 +4y−2=0 ...
$$\sqrt{{x}+\mathrm{3}}\:+\:\sqrt[{\mathrm{3}}]{{x}+\mathrm{1}}\:=\mathrm{2} \\ $$$$\sqrt[{\mathrm{3}}]{{x}+\mathrm{1}}\:={y}\Rightarrow{x}={y}^{\mathrm{3}} −\mathrm{1} \\ $$$$\sqrt{{y}^{\mathrm{3}} +\mathrm{2}}\:=\mathrm{2}−{y} \\ $$$${y}^{\mathrm{3}} +\mathrm{2}={y}^{\mathrm{2}} +\mathrm{4}−\mathrm{4}{y} \\ $$$${y}^{\mathrm{3}} −{y}^{\mathrm{2}} +\mathrm{4}{y}−\mathrm{2}=\mathrm{0} \\ $$$$… \\ $$
Answered by Frix last updated on 27/Mar/25
((x+1))^(1/3) =2−(√(x+3)) x+1=6x+26−(x+15)(√(x+3)) (x+15)(√(x+3))=5x+25 (x+15)^2 (x+3)=(5x+25)^2 x^3 +8x^2 +65x+50=0 x=−(1/3)(8−((1153+150(√(159))))^(1/3) +((−1153+150(√(159))))^(1/3) ) ...better use a calculator to approximate x≈−.848429802
$$\sqrt[{\mathrm{3}}]{{x}+\mathrm{1}}=\mathrm{2}−\sqrt{{x}+\mathrm{3}} \\ $$$${x}+\mathrm{1}=\mathrm{6}{x}+\mathrm{26}−\left({x}+\mathrm{15}\right)\sqrt{{x}+\mathrm{3}} \\ $$$$\left({x}+\mathrm{15}\right)\sqrt{{x}+\mathrm{3}}=\mathrm{5}{x}+\mathrm{25} \\ $$$$\left({x}+\mathrm{15}\right)^{\mathrm{2}} \left({x}+\mathrm{3}\right)=\left(\mathrm{5}{x}+\mathrm{25}\right)^{\mathrm{2}} \\ $$$${x}^{\mathrm{3}} +\mathrm{8}{x}^{\mathrm{2}} +\mathrm{65}{x}+\mathrm{50}=\mathrm{0} \\ $$$${x}=−\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{8}−\sqrt[{\mathrm{3}}]{\mathrm{1153}+\mathrm{150}\sqrt{\mathrm{159}}}+\sqrt[{\mathrm{3}}]{−\mathrm{1153}+\mathrm{150}\sqrt{\mathrm{159}}}\right) \\ $$$$…\mathrm{better}\:\mathrm{use}\:\mathrm{a}\:\mathrm{calculator}\:\mathrm{to}\:\mathrm{approximate} \\ $$$${x}\approx−.\mathrm{848429802} \\ $$
Answered by vnm last updated on 27/Mar/25
x+3=u^2 u+((u^2 −2))^(1/3) =2 u^2 −2=(2−u)^3 u^2 −2=8−12u+6u^2 −u^3 u^3 −5u^2 +12u−10=0 u=t+(5/3) t^3 +((11)/3)t+((20)/(27))=0 t=((−((10)/(27))+(√((53)/(27)))))^(1/3) +((−((10)/(27))−(√((53)/(27)))))^(1/3) x=(t+(5/3))^2 −3
$${x}+\mathrm{3}={u}^{\mathrm{2}} \\ $$$${u}+\sqrt[{\mathrm{3}}]{{u}^{\mathrm{2}} −\mathrm{2}}=\mathrm{2} \\ $$$${u}^{\mathrm{2}} −\mathrm{2}=\left(\mathrm{2}−{u}\right)^{\mathrm{3}} \\ $$$${u}^{\mathrm{2}} −\mathrm{2}=\mathrm{8}−\mathrm{12}{u}+\mathrm{6}{u}^{\mathrm{2}} −{u}^{\mathrm{3}} \\ $$$${u}^{\mathrm{3}} −\mathrm{5}{u}^{\mathrm{2}} +\mathrm{12}{u}−\mathrm{10}=\mathrm{0} \\ $$$${u}={t}+\frac{\mathrm{5}}{\mathrm{3}} \\ $$$${t}^{\mathrm{3}} +\frac{\mathrm{11}}{\mathrm{3}}{t}+\frac{\mathrm{20}}{\mathrm{27}}=\mathrm{0} \\ $$$${t}=\sqrt[{\mathrm{3}}]{−\frac{\mathrm{10}}{\mathrm{27}}+\sqrt{\frac{\mathrm{53}}{\mathrm{27}}}}+\sqrt[{\mathrm{3}}]{−\frac{\mathrm{10}}{\mathrm{27}}−\sqrt{\frac{\mathrm{53}}{\mathrm{27}}}} \\ $$$${x}=\left({t}+\frac{\mathrm{5}}{\mathrm{3}}\right)^{\mathrm{2}} −\mathrm{3} \\ $$

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