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32x-3-48x-2-22x-3-0-

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Question Number 63602 by pavithra last updated on 06/Jul/19
32x^3 −48x^2 −22x−3=0
$$\mathrm{32}{x}^{\mathrm{3}} −\mathrm{48}{x}^{\mathrm{2}} −\mathrm{22}{x}−\mathrm{3}=\mathrm{0} \\ $$
Answered by MJS last updated on 06/Jul/19
x^3 −(3/2)x^2 −((11)/(16))x−(3/(32))=0 x=z+(1/2) z^3 −((23)/(16))z−((11)/(16))=0 D=(1/(27))(−((23)/(16)))^3 +(1/4)(−((11)/(16)))^2 >0 u=((((11)/(32))+((√(2703))/(576))))^(1/3) v=((((11)/(32))−((√(2703))/(576))))^(1/3) x_1 =(1/2)+u+v x_2 =(1/2)+(−(1/2)−((√3)/2)i)u+(−(1/2)+((√3)/2)i)v x_3 =(1/2)+(−(1/2)+((√3)/2)i)u+(−(1/2)−((√3)/2)i)v x_1 ≈1.89000138 x_2 ≈−.195000688−.107600511i x_3 ≈−.195000688+.107600511i
$${x}^{\mathrm{3}} −\frac{\mathrm{3}}{\mathrm{2}}{x}^{\mathrm{2}} −\frac{\mathrm{11}}{\mathrm{16}}{x}−\frac{\mathrm{3}}{\mathrm{32}}=\mathrm{0} \\ $$$${x}={z}+\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${z}^{\mathrm{3}} −\frac{\mathrm{23}}{\mathrm{16}}{z}−\frac{\mathrm{11}}{\mathrm{16}}=\mathrm{0} \\ $$$${D}=\frac{\mathrm{1}}{\mathrm{27}}\left(−\frac{\mathrm{23}}{\mathrm{16}}\right)^{\mathrm{3}} +\frac{\mathrm{1}}{\mathrm{4}}\left(−\frac{\mathrm{11}}{\mathrm{16}}\right)^{\mathrm{2}} >\mathrm{0} \\ $$$${u}=\sqrt[{\mathrm{3}}]{\frac{\mathrm{11}}{\mathrm{32}}+\frac{\sqrt{\mathrm{2703}}}{\mathrm{576}}}\:\:\:\:\:{v}=\sqrt[{\mathrm{3}}]{\frac{\mathrm{11}}{\mathrm{32}}−\frac{\sqrt{\mathrm{2703}}}{\mathrm{576}}} \\ $$$${x}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}+{u}+{v} \\ $$$${x}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}+\left(−\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}\right){u}+\left(−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}\right){v} \\ $$$${x}_{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{2}}+\left(−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}\right){u}+\left(−\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}\right){v} \\ $$$${x}_{\mathrm{1}} \approx\mathrm{1}.\mathrm{89000138} \\ $$$${x}_{\mathrm{2}} \approx−.\mathrm{195000688}−.\mathrm{107600511i} \\ $$$${x}_{\mathrm{3}} \approx−.\mathrm{195000688}+.\mathrm{107600511i} \\ $$
Commented by pavithra last updated on 07/Jul/19
thankyou
$${thankyou} \\ $$
Answered by ajfour last updated on 07/Jul/19
let x=((at+b)/(t+1)) 32(a^3 t^3 +3a^2 bt^2 +3ab^2 t+b^3 ) −48(a^2 t^2 +2abt+b^2 )(t+1) −22(at+b)(t^2 +2t+1) −3(t^3 +3t^2 +3t+1)=0 ⇒ (32a^3 −48a^2 −22a−3)t^3 +(96a^2 b−48a^2 −96ab−44a−22b−9)t^2 +(96ab^2 −96ab−48b^2 −22a−44b−9)t +32b^3 −48b^2 −22b−3=0 let coefficients of t^2 and t be zero; ⇒ 96a^2 b−48a^2 −96ab−44a−22b−9=0 96ab^2 −96ab−48b^2 −22a−44b−9=0 subtracting , knowing a≠b 48ab−24(a+b)−11=0 ...(i) adding 48ab(a+b)−24(a+b)^2 −48ab −33(a+b)−9=0 using (i) 24(a+b)^2 +11(a+b)−24(a+b)^2 −24(a+b)−11−33(a+b)−9=0 ⇒ a+b=−((10)/(23)) 48ab=−((240)/(23))+11=((13)/(23)) ⇒ ab=((13)/(23×48)) a,b are roots of eq. x^2 +((10x)/(23))+((13)/(23×48))=0 a,b =−(5/(23))±(√(((5/(23)))^2 −((13)/(23×16)))) a,b =((−5±(√(101)))/(23)) t^3 =((32b^3 −48b^2 −22b−3)/(32a^3 −48a^2 −22a−3)) .....
$${let}\:\:{x}=\frac{{at}+{b}}{{t}+\mathrm{1}} \\ $$$$\mathrm{32}\left({a}^{\mathrm{3}} {t}^{\mathrm{3}} +\mathrm{3}{a}^{\mathrm{2}} {bt}^{\mathrm{2}} +\mathrm{3}{ab}^{\mathrm{2}} {t}+{b}^{\mathrm{3}} \right) \\ $$$$−\mathrm{48}\left({a}^{\mathrm{2}} {t}^{\mathrm{2}} +\mathrm{2}{abt}+{b}^{\mathrm{2}} \right)\left({t}+\mathrm{1}\right) \\ $$$$−\mathrm{22}\left({at}+{b}\right)\left({t}^{\mathrm{2}} +\mathrm{2}{t}+\mathrm{1}\right) \\ $$$$−\mathrm{3}\left({t}^{\mathrm{3}} +\mathrm{3}{t}^{\mathrm{2}} +\mathrm{3}{t}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow \\ $$$$\left(\mathrm{32}{a}^{\mathrm{3}} −\mathrm{48}{a}^{\mathrm{2}} −\mathrm{22}{a}−\mathrm{3}\right){t}^{\mathrm{3}} \\ $$$$+\left(\mathrm{96}{a}^{\mathrm{2}} {b}−\mathrm{48}{a}^{\mathrm{2}} −\mathrm{96}{ab}−\mathrm{44}{a}−\mathrm{22}{b}−\mathrm{9}\right){t}^{\mathrm{2}} \\ $$$$+\left(\mathrm{96}{ab}^{\mathrm{2}} −\mathrm{96}{ab}−\mathrm{48}{b}^{\mathrm{2}} −\mathrm{22}{a}−\mathrm{44}{b}−\mathrm{9}\right){t} \\ $$$$+\mathrm{32}{b}^{\mathrm{3}} −\mathrm{48}{b}^{\mathrm{2}} −\mathrm{22}{b}−\mathrm{3}=\mathrm{0} \\ $$$${let}\:\:{coefficients}\:{of}\:{t}^{\mathrm{2}} \:{and}\:{t}\:{be}\:{zero}; \\ $$$$\Rightarrow \\ $$$$\mathrm{96}{a}^{\mathrm{2}} {b}−\mathrm{48}{a}^{\mathrm{2}} −\mathrm{96}{ab}−\mathrm{44}{a}−\mathrm{22}{b}−\mathrm{9}=\mathrm{0} \\ $$$$\mathrm{96}{ab}^{\mathrm{2}} −\mathrm{96}{ab}−\mathrm{48}{b}^{\mathrm{2}} −\mathrm{22}{a}−\mathrm{44}{b}−\mathrm{9}=\mathrm{0} \\ $$$${subtracting}\:,\:{knowing}\:{a}\neq{b} \\ $$$$\mathrm{48}{ab}−\mathrm{24}\left({a}+{b}\right)−\mathrm{11}=\mathrm{0}\:\:\:\:\:\:…\left({i}\right) \\ $$$${adding} \\ $$$$\mathrm{48}{ab}\left({a}+{b}\right)−\mathrm{24}\left({a}+{b}\right)^{\mathrm{2}} −\mathrm{48}{ab} \\ $$$$\:\:\:\:−\mathrm{33}\left({a}+{b}\right)−\mathrm{9}=\mathrm{0} \\ $$$${using}\:\left({i}\right) \\ $$$$\mathrm{24}\left({a}+{b}\right)^{\mathrm{2}} +\mathrm{11}\left({a}+{b}\right)−\mathrm{24}\left({a}+{b}\right)^{\mathrm{2}} \\ $$$$−\mathrm{24}\left({a}+{b}\right)−\mathrm{11}−\mathrm{33}\left({a}+{b}\right)−\mathrm{9}=\mathrm{0} \\ $$$$\Rightarrow\:\:{a}+{b}=−\frac{\mathrm{10}}{\mathrm{23}} \\ $$$$\mathrm{48}{ab}=−\frac{\mathrm{240}}{\mathrm{23}}+\mathrm{11}=\frac{\mathrm{13}}{\mathrm{23}} \\ $$$$\Rightarrow\:{ab}=\frac{\mathrm{13}}{\mathrm{23}×\mathrm{48}} \\ $$$${a},{b}\:\:{are}\:{roots}\:{of}\:{eq}. \\ $$$$\:\:\:{x}^{\mathrm{2}} +\frac{\mathrm{10}{x}}{\mathrm{23}}+\frac{\mathrm{13}}{\mathrm{23}×\mathrm{48}}=\mathrm{0} \\ $$$${a},{b}\:=−\frac{\mathrm{5}}{\mathrm{23}}\pm\sqrt{\left(\frac{\mathrm{5}}{\mathrm{23}}\right)^{\mathrm{2}} −\frac{\mathrm{13}}{\mathrm{23}×\mathrm{16}}} \\ $$$$\:{a},{b}\:=\frac{−\mathrm{5}\pm\sqrt{\mathrm{101}}}{\mathrm{23}} \\ $$$$\:\:\:\:{t}^{\mathrm{3}} =\frac{\mathrm{32}{b}^{\mathrm{3}} −\mathrm{48}{b}^{\mathrm{2}} −\mathrm{22}{b}−\mathrm{3}}{\mathrm{32}{a}^{\mathrm{3}} −\mathrm{48}{a}^{\mathrm{2}} −\mathrm{22}{a}−\mathrm{3}} \\ $$$$….. \\ $$

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