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Question Number 178785 by Mastermind last updated on 21/Oct/22

Answered by Rasheed.Sindhi last updated on 21/Oct/22

  (((3−x)/(3+x)))^(1/4)  −(√((3+x)/(3−x))) =(√2)  Let (((3−x)/(3+x)))^(1/4)  =y          (√((3−x)/(3+x))) =y^2 ⇒(√((3+x)/(3−x))) =(1/y^2 )  y−(1/y^2 )=(√2)  y^3 −(√2) y^2 −1=0  This I can′t solve.  ....

$$ \\ $$$$\sqrt[{\mathrm{4}}]{\frac{\mathrm{3}−{x}}{\mathrm{3}+{x}}}\:−\sqrt{\frac{\mathrm{3}+{x}}{\mathrm{3}−{x}}}\:=\sqrt{\mathrm{2}} \\ $$$${Let}\:\sqrt[{\mathrm{4}}]{\frac{\mathrm{3}−{x}}{\mathrm{3}+{x}}}\:={y} \\ $$$$\:\:\:\:\:\:\:\:\sqrt{\frac{\mathrm{3}−{x}}{\mathrm{3}+{x}}}\:={y}^{\mathrm{2}} \Rightarrow\sqrt{\frac{\mathrm{3}+{x}}{\mathrm{3}−{x}}}\:=\frac{\mathrm{1}}{{y}^{\mathrm{2}} } \\ $$$${y}−\frac{\mathrm{1}}{{y}^{\mathrm{2}} }=\sqrt{\mathrm{2}} \\ $$$${y}^{\mathrm{3}} −\sqrt{\mathrm{2}}\:{y}^{\mathrm{2}} −\mathrm{1}=\mathrm{0} \\ $$$$\mathcal{T}{his}\:{I}\:{can}'{t}\:{solve}. \\ $$$$.... \\ $$

Commented by cortano1 last updated on 21/Oct/22

typo . it′s (((3−x)/(3+x)))^(1/4)

$$\mathrm{typo}\:.\:\mathrm{it}'\mathrm{s}\:\sqrt[{\mathrm{4}}]{\frac{\mathrm{3}−\mathrm{x}}{\mathrm{3}+\mathrm{x}}} \\ $$

Commented by Rasheed.Sindhi last updated on 21/Oct/22

ThanX sir, corrected!

$$\mathcal{T}{han}\mathcal{X}\:{sir},\:{corrected}! \\ $$

Commented by mr W last updated on 21/Oct/22

y^3 −(√2)y^2 −1=0  (1/y^3 )+((√2)/y)−1=0  (1/y)=(((√((1/4)+((2(√2))/(27))))+(1/2)))^(1/3) −(((√((1/4)+((2(√2))/(27))))−(1/2)))^(1/3)   ⇒y=(1/( (((√((1/4)+((2(√2))/(27))))+(1/2)))^(1/3) −(((√((1/4)+((2(√2))/(27))))−(1/2)))^(1/3) ))

$${y}^{\mathrm{3}} −\sqrt{\mathrm{2}}{y}^{\mathrm{2}} −\mathrm{1}=\mathrm{0} \\ $$$$\frac{\mathrm{1}}{{y}^{\mathrm{3}} }+\frac{\sqrt{\mathrm{2}}}{{y}}−\mathrm{1}=\mathrm{0} \\ $$$$\frac{\mathrm{1}}{{y}}=\sqrt[{\mathrm{3}}]{\sqrt{\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{27}}}+\frac{\mathrm{1}}{\mathrm{2}}}−\sqrt[{\mathrm{3}}]{\sqrt{\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{27}}}−\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\Rightarrow{y}=\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\sqrt{\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{27}}}+\frac{\mathrm{1}}{\mathrm{2}}}−\sqrt[{\mathrm{3}}]{\sqrt{\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{27}}}−\frac{\mathrm{1}}{\mathrm{2}}}} \\ $$

Commented by Spillover last updated on 21/Oct/22

thank you

$$\mathrm{thank}\:\mathrm{you} \\ $$

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