Question Number 8032 by Nayon last updated on 28/Sep/16
$${find}\:{the}\:{real}\:{root}: \\ $$$$\mathrm{99}{x}^{\mathrm{3}} +\mathrm{297}{x}^{\mathrm{2}} +\mathrm{594}{x}−\mathrm{7867}=\mathrm{0} \\ $$
Answered by prakash jain last updated on 28/Sep/16
![x=y−1 for solving cubic ax^3 +bx^2 +cx+d=0first step is to put x=y−(b/(3a)) (to remove quadratic term) 99(y−1)^3 +297(y−1)^2 +594(y−1)−7867= 99(y^3 −3y^2 +3y−1)+297(y^2 −2y+1) +594y−594−7867=0 99y^3 +297y−99−2×297y+297 +594y−8461=0 99y^3 +297y−8263=0 99y^3 +297y=8263 y^3 +3y=((8263)/(99)) A solution of equation y^3 +Ay=B is s−t where s and t satisfy 3st=A s^3 −t^3 =B [∵(s−t)^3 +3st(s−t)=s^3 −t^3 ] 3st=3⇒st=1 s^3 −t^3 =((8263)/(99)) (1/t^3 )−t^3 =((8263)/(99)) t^3 =u 1−u^2 =((8263)/(99))u u^2 +((8263)/(99))u−1=0 t^3 =u=((−((8263)/(99))±(√((((8263)/(99)))^2 +4)))/2) =(((√(68316373))−8263)/(198)) (taking only +ve root) s^3 −t^3 =((8263)/(99))⇒s^3 =((8263)/(99))+(((√(68316373))−8293)/(198)) =((8263+(√(68316373)))/(198)) y=s−t=(((8263+(√(68316373)))/(198)))^(1/3) −(((−8263+(√(68316373)))/(198)))^(1/3) =4.3704−0.2288=4.1416 x=y−1≈3.1416](https://www.tinkutara.com/question/Q8051.png)
$${x}={y}−\mathrm{1} \\ $$$$\mathrm{for}\:\mathrm{solving}\:\mathrm{cubic}\:{ax}^{\mathrm{3}} +{bx}^{\mathrm{2}} +{cx}+{d}=\mathrm{0first}\:\mathrm{step} \\ $$$$\mathrm{is}\:\mathrm{to}\:\mathrm{put}\:{x}={y}−\frac{{b}}{\mathrm{3}{a}}\:\left({to}\:{remove}\:{quadratic}\:{term}\right) \\ $$$$\mathrm{99}\left({y}−\mathrm{1}\right)^{\mathrm{3}} +\mathrm{297}\left({y}−\mathrm{1}\overset{\mathrm{2}} {\right)}+\mathrm{594}\left({y}−\mathrm{1}\right)−\mathrm{7867}= \\ $$$$\mathrm{99}\left({y}^{\mathrm{3}} −\mathrm{3}{y}^{\mathrm{2}} +\mathrm{3}{y}−\mathrm{1}\right)+\mathrm{297}\left({y}^{\mathrm{2}} −\mathrm{2}{y}+\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{594}{y}−\mathrm{594}−\mathrm{7867}=\mathrm{0} \\ $$$$\mathrm{99}{y}^{\mathrm{3}} +\mathrm{297}{y}−\mathrm{99}−\mathrm{2}×\mathrm{297}{y}+\mathrm{297} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:+\mathrm{594}{y}−\mathrm{8461}=\mathrm{0} \\ $$$$\mathrm{99}{y}^{\mathrm{3}} +\mathrm{297}{y}−\mathrm{8263}=\mathrm{0} \\ $$$$\mathrm{99}{y}^{\mathrm{3}} +\mathrm{297}{y}=\mathrm{8263} \\ $$$${y}^{\mathrm{3}} +\mathrm{3}{y}=\frac{\mathrm{8263}}{\mathrm{99}} \\ $$$$\mathrm{A}\:\mathrm{solution}\:\mathrm{of}\:\mathrm{equation}\:{y}^{\mathrm{3}} +{Ay}={B} \\ $$$$\mathrm{is}\:{s}−{t}\:{where}\:{s}\:{and}\:{t}\:{satisfy} \\ $$$$\mathrm{3}{st}={A}\: \\ $$$${s}^{\mathrm{3}} −{t}^{\mathrm{3}} ={B}\:\left[\because\left({s}−{t}\right)^{\mathrm{3}} +\mathrm{3}{st}\left({s}−{t}\right)={s}^{\mathrm{3}} −{t}^{\mathrm{3}} \right] \\ $$$$\mathrm{3}{st}=\mathrm{3}\Rightarrow{st}=\mathrm{1} \\ $$$${s}^{\mathrm{3}} −{t}^{\mathrm{3}} =\frac{\mathrm{8263}}{\mathrm{99}} \\ $$$$\frac{\mathrm{1}}{{t}^{\mathrm{3}} }−{t}^{\mathrm{3}} =\frac{\mathrm{8263}}{\mathrm{99}} \\ $$$${t}^{\mathrm{3}} ={u} \\ $$$$\mathrm{1}−{u}^{\mathrm{2}} =\frac{\mathrm{8263}}{\mathrm{99}}{u} \\ $$$${u}^{\mathrm{2}} +\frac{\mathrm{8263}}{\mathrm{99}}{u}−\mathrm{1}=\mathrm{0} \\ $$$${t}^{\mathrm{3}} ={u}=\frac{−\frac{\mathrm{8263}}{\mathrm{99}}\pm\sqrt{\left(\frac{\mathrm{8263}}{\mathrm{99}}\right)^{\mathrm{2}} +\mathrm{4}}}{\mathrm{2}} \\ $$$$=\frac{\sqrt{\mathrm{68316373}}−\mathrm{8263}}{\mathrm{198}}\:\left(\mathrm{taking}\:\mathrm{only}\:+\mathrm{ve}\:\mathrm{root}\right) \\ $$$${s}^{\mathrm{3}} −{t}^{\mathrm{3}} =\frac{\mathrm{8263}}{\mathrm{99}}\Rightarrow{s}^{\mathrm{3}} =\frac{\mathrm{8263}}{\mathrm{99}}+\frac{\sqrt{\mathrm{68316373}}−\mathrm{8293}}{\mathrm{198}} \\ $$$$=\frac{\mathrm{8263}+\sqrt{\mathrm{68316373}}}{\mathrm{198}} \\ $$$${y}={s}−{t}=\left(\frac{\mathrm{8263}+\sqrt{\mathrm{68316373}}}{\mathrm{198}}\right)^{\mathrm{1}/\mathrm{3}} −\left(\frac{−\mathrm{8263}+\sqrt{\mathrm{68316373}}}{\mathrm{198}}\right)^{\mathrm{1}/\mathrm{3}} \\ $$$$=\mathrm{4}.\mathrm{3704}−\mathrm{0}.\mathrm{2288}=\mathrm{4}.\mathrm{1416} \\ $$$${x}={y}−\mathrm{1}\approx\mathrm{3}.\mathrm{1416} \\ $$
Commented by prakash jain last updated on 28/Sep/16
$$\mathrm{Is}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{derived}\:\mathrm{from}\:\mathrm{some} \\ $$$$\mathrm{approximate}\:\mathrm{value}\:\mathrm{of}\:\pi? \\ $$