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v-2b-3cp-p-3-bp-2-3v-2-2bpv-3cp-b-1-bp-2-cp-3-gt-0-b-c-R-b-lt-0-Any-non-zero-real-value-of-p-in-terms-of-b-c-obeying-above-condition-




Question Number 80881 by ajfour last updated on 07/Feb/20
v=−(((2b+3cp)p)/(3+bp^2 ))   ((3v^2 +2bpv+3cp+b)/(1+bp^2 +cp^3 )) > 0    b,c ∈ R , b<0  Any non-zero real value of p  in terms of b,c  obeying above  condition?
$${v}=−\frac{\left(\mathrm{2}{b}+\mathrm{3}{cp}\right){p}}{\mathrm{3}+{bp}^{\mathrm{2}} } \\ $$$$\:\frac{\mathrm{3}{v}^{\mathrm{2}} +\mathrm{2}{bpv}+\mathrm{3}{cp}+{b}}{\mathrm{1}+{bp}^{\mathrm{2}} +{cp}^{\mathrm{3}} }\:>\:\mathrm{0}\:\: \\ $$$${b},{c}\:\in\:\mathbb{R}\:,\:{b}<\mathrm{0} \\ $$$${Any}\:{non}-{zero}\:{real}\:{value}\:{of}\:{p} \\ $$$${in}\:{terms}\:{of}\:{b},{c}\:\:{obeying}\:{above} \\ $$$${condition}? \\ $$
Answered by ajfour last updated on 07/Feb/20
let  p=−(b/c)  v=((−b^2 )/(c(3+(b^3 /c^2 ))))=−((b^2 c)/(3c^2 +b^3 ))  S=(((b^2 c)/(3c^2 +b^3 )))^2 +((2b^2 )/c)(((b^2 c)/(3c^2 +b^3 )))−2b  let me test it  x^3 −49x+120=0  let x=((u+v)/(1+pu))  u^3 +Su+((v^3 +bv+c)/(1+bp^2 +cp^3 ))=0  ⇒ u^3 +Su+T=0  D= (T^(  2) /4)+(S^3 /(27))  u={−(T/2)+(√D)}^(1/3) +{−(T/2)−(√D)}^(1/3)   ........
$${let}\:\:{p}=−\frac{{b}}{{c}} \\ $$$${v}=\frac{−{b}^{\mathrm{2}} }{{c}\left(\mathrm{3}+\frac{{b}^{\mathrm{3}} }{{c}^{\mathrm{2}} }\right)}=−\frac{{b}^{\mathrm{2}} {c}}{\mathrm{3}{c}^{\mathrm{2}} +{b}^{\mathrm{3}} } \\ $$$${S}=\left(\frac{{b}^{\mathrm{2}} {c}}{\mathrm{3}{c}^{\mathrm{2}} +{b}^{\mathrm{3}} }\right)^{\mathrm{2}} +\frac{\mathrm{2}{b}^{\mathrm{2}} }{{c}}\left(\frac{{b}^{\mathrm{2}} {c}}{\mathrm{3}{c}^{\mathrm{2}} +{b}^{\mathrm{3}} }\right)−\mathrm{2}{b} \\ $$$${let}\:{me}\:{test}\:{it} \\ $$$${x}^{\mathrm{3}} −\mathrm{49}{x}+\mathrm{120}=\mathrm{0} \\ $$$${let}\:{x}=\frac{{u}+{v}}{\mathrm{1}+{pu}} \\ $$$${u}^{\mathrm{3}} +{Su}+\frac{{v}^{\mathrm{3}} +{bv}+{c}}{\mathrm{1}+{bp}^{\mathrm{2}} +{cp}^{\mathrm{3}} }=\mathrm{0} \\ $$$$\Rightarrow\:{u}^{\mathrm{3}} +{Su}+{T}=\mathrm{0} \\ $$$${D}=\:\frac{{T}^{\:\:\mathrm{2}} }{\mathrm{4}}+\frac{{S}^{\mathrm{3}} }{\mathrm{27}} \\ $$$${u}=\left\{−\frac{{T}}{\mathrm{2}}+\sqrt{{D}}\right\}^{\mathrm{1}/\mathrm{3}} +\left\{−\frac{{T}}{\mathrm{2}}−\sqrt{{D}}\right\}^{\mathrm{1}/\mathrm{3}} \\ $$$$…….. \\ $$

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