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Question Number 198474 by cortano12 last updated on 20/Oct/23
16000 = (x^3 /((1−x)^2 )) x=?
$$\mathrm{16000}\:=\:\frac{\mathrm{x}^{\mathrm{3}} }{\left(\mathrm{1}−\mathrm{x}\right)^{\mathrm{2}} }\: \\ $$$$\:\mathrm{x}=? \\ $$
Answered by Frix last updated on 20/Oct/23
n=(x^3 /((1−x)^2 )); n>((27)/4) ⇒ x_1 =(a/3)−((2(√(a(a−6))))/3)cos ((2π+sin^(−1) ((2a^2 −18a+27)/(2(√(a(a−6)^3 )))))/3) x_2 =(a/3)−((2(√(a(a−6))))/3)sin ((sin^(−1) ((2a^2 −18a+27)/(2(√(a(a−6)^3 )))))/3) x_3 =(a/3)+((2(√(a(a−6))))/3)sin ((π+sin^(−1) ((2a^2 −18a+27)/(2(√(a(a−6)^3 )))))/3) With a=16000 x_1 ≈.992160710 x_2 ≈1.00802683 x_3 ≈15997.9998
$${n}=\frac{{x}^{\mathrm{3}} }{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} };\:{n}>\frac{\mathrm{27}}{\mathrm{4}}\:\Rightarrow \\ $$$${x}_{\mathrm{1}} =\frac{{a}}{\mathrm{3}}−\frac{\mathrm{2}\sqrt{{a}\left({a}−\mathrm{6}\right)}}{\mathrm{3}}\mathrm{cos}\:\frac{\mathrm{2}\pi+\mathrm{sin}^{−\mathrm{1}} \:\frac{\mathrm{2}{a}^{\mathrm{2}} −\mathrm{18}{a}+\mathrm{27}}{\mathrm{2}\sqrt{{a}\left({a}−\mathrm{6}\right)^{\mathrm{3}} }}}{\mathrm{3}} \\ $$$${x}_{\mathrm{2}} =\frac{{a}}{\mathrm{3}}−\frac{\mathrm{2}\sqrt{{a}\left({a}−\mathrm{6}\right)}}{\mathrm{3}}\mathrm{sin}\:\frac{\mathrm{sin}^{−\mathrm{1}} \:\frac{\mathrm{2}{a}^{\mathrm{2}} −\mathrm{18}{a}+\mathrm{27}}{\mathrm{2}\sqrt{{a}\left({a}−\mathrm{6}\right)^{\mathrm{3}} }}}{\mathrm{3}} \\ $$$${x}_{\mathrm{3}} =\frac{{a}}{\mathrm{3}}+\frac{\mathrm{2}\sqrt{{a}\left({a}−\mathrm{6}\right)}}{\mathrm{3}}\mathrm{sin}\:\frac{\pi+\mathrm{sin}^{−\mathrm{1}} \:\frac{\mathrm{2}{a}^{\mathrm{2}} −\mathrm{18}{a}+\mathrm{27}}{\mathrm{2}\sqrt{{a}\left({a}−\mathrm{6}\right)^{\mathrm{3}} }}}{\mathrm{3}} \\ $$$$\mathrm{With}\:{a}=\mathrm{16000} \\ $$$${x}_{\mathrm{1}} \approx.\mathrm{992160710} \\ $$$${x}_{\mathrm{2}} \approx\mathrm{1}.\mathrm{00802683} \\ $$$${x}_{\mathrm{3}} \approx\mathrm{15997}.\mathrm{9998} \\ $$
Answered by a.lgnaoui last updated on 20/Oct/23
4^2 (1−x)^2 =((x/(10)))^3 40(√(10)) (1−x)=x(√x) x(√x) +40(√(10)) x−40(√(10)) =0 posons z=(√x) z^3 +40(√(10)) z^2 −40(√(10)) =0 ⇒z_1 =−126,483199724 z_2 =−1,003924108 z_3 =0,99608572826 soit x_1 = 0,99218677 x_2 = 1,00800076 x_3 =15997,99981242
$$\mathrm{4}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{x}\right)^{\mathrm{2}} =\left(\frac{\mathrm{x}}{\mathrm{10}}\right)^{\mathrm{3}} \\ $$$$\mathrm{40}\sqrt{\mathrm{10}}\:\left(\mathrm{1}−\mathrm{x}\right)=\mathrm{x}\sqrt{\mathrm{x}} \\ $$$$\mathrm{x}\sqrt{\mathrm{x}}\:+\mathrm{40}\sqrt{\mathrm{10}}\:\mathrm{x}−\mathrm{40}\sqrt{\mathrm{10}}\:=\mathrm{0} \\ $$$$\mathrm{posons}\:\:\:\boldsymbol{\mathrm{z}}=\sqrt{\boldsymbol{\mathrm{x}}}\: \\ $$$$ \\ $$$$\boldsymbol{\mathrm{z}}^{\mathrm{3}} +\mathrm{40}\sqrt{\mathrm{10}}\:\boldsymbol{\mathrm{z}}^{\mathrm{2}} −\mathrm{40}\sqrt{\mathrm{10}}\:=\mathrm{0} \\ $$$$ \\ $$$$\Rightarrow\boldsymbol{\mathrm{z}}_{\mathrm{1}} =−\mathrm{126},\mathrm{483199724}\:\: \\ $$$$\:\:\:\:\:\:\boldsymbol{\mathrm{z}}_{\mathrm{2}} =−\mathrm{1},\mathrm{003924108} \\ $$$$\:\:\:\:\:\:\mathrm{z}_{\mathrm{3}} =\mathrm{0},\mathrm{99608572826} \\ $$$$\mathrm{soit}\:\boldsymbol{\mathrm{x}}_{\mathrm{1}} =\:\:\:\:\:\:\:\:\:\:\mathrm{0},\mathrm{99218677} \\ $$$$\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{x}}_{\mathrm{2}} =\:\:\:\:\:\:\:\:\:\:\mathrm{1},\mathrm{00800076} \\ $$$$\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{x}}_{\mathrm{3}} =\mathrm{15997},\mathrm{99981242} \\ $$$$\:\:\:\:\:\:\: \\ $$

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