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if-x-4-x-3-0-S-amp-P-are-amp-real-roots-what-is-value-of-4P-2-S-2-

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Question Number 202873 by lorance last updated on 05/Jan/24
if x^4 −x−3=0 S & P are + & × real roots what is value of 4P^2 +S^2 ?
$${if}\:{x}^{\mathrm{4}} −{x}−\mathrm{3}=\mathrm{0} \\ $$$$\mathrm{S}\:\&\:\mathrm{P}\:{are}\:+\:\&\:×\:{real}\:{roots} \\ $$$${what}\:{is}\:{value}\:{of}\:\mathrm{4}{P}^{\mathrm{2}} \:+{S}^{\mathrm{2}} ? \\ $$
Commented by Frix last updated on 05/Jan/24
x^4 +px+q=0 x_1 =a x_2 =b x_(3, 4) =−((a+b)/2)±((√(3a^2 +2ab+3b^2 ))/2)i ⇒ x^4 −(a+b)(a^2 +b^2 )x+a(a^2 +ab+b^2 )b=0 p=−(a+b)(a^2 +b^2 )∧q=a(a^2 +ab+b^2 )b You ask for 4a^2 b^2 +(a+b)^2 =r Let u=a+b∧v=ab p=2uv−u^3 q=u^2 v−v^2 This doesn′t lead to any nice solution for r=u^2 +4v^2 generally and also not with the given p=−1∧q=−3: 2uv−u^3 =−1 ⇒ v=((u^3 −1)/(2u)) u^2 v−v^2 =−3 ⇒ u^6 +12u^3 −1=0 ⇒ u=±(√(((((√(257))/2)+(1/2)))^(1/3) −((((√(257))/2)−(1/2)))^(1/3) )) r=u^4 +u^2 −2u+(1/u^2 ) No chance for something useable...
$${x}^{\mathrm{4}} +{px}+{q}=\mathrm{0} \\ $$$${x}_{\mathrm{1}} ={a} \\ $$$${x}_{\mathrm{2}} ={b} \\ $$$${x}_{\mathrm{3},\:\mathrm{4}} =−\frac{{a}+{b}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{3}{a}^{\mathrm{2}} +\mathrm{2}{ab}+\mathrm{3}{b}^{\mathrm{2}} }}{\mathrm{2}}\mathrm{i} \\ $$$$\Rightarrow \\ $$$${x}^{\mathrm{4}} −\left({a}+{b}\right)\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right){x}+{a}\left({a}^{\mathrm{2}} +{ab}+{b}^{\mathrm{2}} \right){b}=\mathrm{0} \\ $$$${p}=−\left({a}+{b}\right)\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\wedge{q}={a}\left({a}^{\mathrm{2}} +{ab}+{b}^{\mathrm{2}} \right){b} \\ $$$$\mathrm{You}\:\mathrm{ask}\:\mathrm{for}\:\mathrm{4}{a}^{\mathrm{2}} {b}^{\mathrm{2}} +\left({a}+{b}\right)^{\mathrm{2}} ={r} \\ $$$$\mathrm{Let}\:{u}={a}+{b}\wedge{v}={ab} \\ $$$${p}=\mathrm{2}{uv}−{u}^{\mathrm{3}} \\ $$$${q}={u}^{\mathrm{2}} {v}−{v}^{\mathrm{2}} \\ $$$$\mathrm{This}\:\mathrm{doesn}'\mathrm{t}\:\mathrm{lead}\:\mathrm{to}\:\mathrm{any}\:\mathrm{nice}\:\mathrm{solution}\:\mathrm{for} \\ $$$${r}={u}^{\mathrm{2}} +\mathrm{4}{v}^{\mathrm{2}} \:\mathrm{generally}\:\mathrm{and}\:\mathrm{also}\:\mathrm{not}\:\mathrm{with}\:\mathrm{the} \\ $$$$\mathrm{given}\:{p}=−\mathrm{1}\wedge{q}=−\mathrm{3}: \\ $$$$\mathrm{2}{uv}−{u}^{\mathrm{3}} =−\mathrm{1}\:\Rightarrow\:{v}=\frac{{u}^{\mathrm{3}} −\mathrm{1}}{\mathrm{2}{u}} \\ $$$${u}^{\mathrm{2}} {v}−{v}^{\mathrm{2}} =−\mathrm{3}\:\Rightarrow\:{u}^{\mathrm{6}} +\mathrm{12}{u}^{\mathrm{3}} −\mathrm{1}=\mathrm{0}\:\Rightarrow \\ $$$${u}=\pm\sqrt{\sqrt[{\mathrm{3}}]{\frac{\sqrt{\mathrm{257}}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}}−\sqrt[{\mathrm{3}}]{\frac{\sqrt{\mathrm{257}}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}}} \\ $$$${r}={u}^{\mathrm{4}} +{u}^{\mathrm{2}} −\mathrm{2}{u}+\frac{\mathrm{1}}{{u}^{\mathrm{2}} } \\ $$$$\mathrm{No}\:\mathrm{chance}\:\mathrm{for}\:\mathrm{something}\:\mathrm{useable}… \\ $$
Answered by Rasheed.Sindhi last updated on 05/Jan/24
let a,b are real and c,d are imaginary roots. (x−a)(x−b)(x−c)(x−d)= x^4 −x−3=0 x^4 −(a+b+c+d)x^3 +(ab+ac+ad+bc+bd+cd)x^2 −(abc+abd+acd+bcd)x+abcd = x^4 −x−3=0 a+b+c+d=0 ∧ ab+ac+ad+bc+bd+cd=0 ∧ abc+abd+acd+bcd=1 ∧ abcd=−3 (a+b+c+d)^2 =0^2 a^2 +b^2 +c^2 +d^2 +2(ab+ac+ad+bc+bd+cd)=0 a^2 +b^2 +c^2 +d^2 =0 ..... ...
$${let}\:{a},{b}\:{are}\:{real}\:{and}\:{c},{d}\:{are}\:{imaginary}\:{roots}. \\ $$$$\left({x}−{a}\right)\left({x}−{b}\right)\left({x}−{c}\right)\left({x}−{d}\right)=\:{x}^{\mathrm{4}} −{x}−\mathrm{3}=\mathrm{0} \\ $$$${x}^{\mathrm{4}} −\left({a}+{b}+{c}+{d}\right){x}^{\mathrm{3}} +\left({ab}+{ac}+{ad}+{bc}+{bd}+{cd}\right){x}^{\mathrm{2}} \\ $$$$\:\:\:\:\:−\left({abc}+{abd}+{acd}+{bcd}\right){x}+{abcd}\:=\:{x}^{\mathrm{4}} −{x}−\mathrm{3}=\mathrm{0} \\ $$$${a}+{b}+{c}+{d}=\mathrm{0}\:\wedge\:{ab}+{ac}+{ad}+{bc}+{bd}+{cd}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\wedge\:{abc}+{abd}+{acd}+{bcd}=\mathrm{1}\:\wedge\:{abcd}=−\mathrm{3} \\ $$$$\left({a}+{b}+{c}+{d}\right)^{\mathrm{2}} =\mathrm{0}^{\mathrm{2}} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{d}^{\mathrm{2}} +\mathrm{2}\left({ab}+{ac}+{ad}+{bc}+{bd}+{cd}\right)=\mathrm{0} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{d}^{\mathrm{2}} =\mathrm{0} \\ $$$$….. \\ $$$$… \\ $$

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