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Find-the-real-zeros-of-the-polynomial-P-a-x-x-2-1-x-1-2-ax-2-where-a-is-a-given-real-number-

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Question Number 152544 by liberty last updated on 29/Aug/21
Find the real zeros of the polynomial P_a (x)=(x^2 +1)(x−1)^2 −ax^2 where a is a given real number
$$\:\mathrm{Find}\:\mathrm{the}\:\mathrm{real}\:\mathrm{zeros}\:\mathrm{of}\:\mathrm{the}\:\mathrm{polynomial} \\ $$$$\:\mathrm{P}_{\mathrm{a}} \left(\mathrm{x}\right)=\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{x}−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{ax}^{\mathrm{2}} \\ $$$$\mathrm{where}\:\mathrm{a}\:\mathrm{is}\:\mathrm{a}\:\mathrm{given}\:\mathrm{real}\:\mathrm{number} \\ $$
Commented by MJS_new last updated on 30/Aug/21
I get x_1 =((1−(√(a+1))−(√(a−2−2(√(a+1)))))/2) ∈R with a≥8 x_2 =((1−(√(a+1))+(√(a−2−2(√(a+1)))))/2) ∈R with a≥8 x_3 =((1+(√(a+1))−(√(a−2+2(√(a+1)))))/2) ∈R with a≥0 x_4 =((1+(√(a+1))+(√(a−2+2(√(a+1)))))/2) ∈R with a≥0 btw (1) x_2 =(1/x_1 )∧x_4 =(1/x_3 ) because P_a (x)=x^4 −2x^3 −(a−2)x^2 −2x+1 symmetric (2) no real solution for a<0 which is easy to see: P_a (x)=0 ⇒ a=(((x^2 +1)(x−1)^2 )/x^2 )≥0
$$\mathrm{I}\:\mathrm{get} \\ $$$${x}_{\mathrm{1}} =\frac{\mathrm{1}−\sqrt{{a}+\mathrm{1}}−\sqrt{{a}−\mathrm{2}−\mathrm{2}\sqrt{{a}+\mathrm{1}}}}{\mathrm{2}}\:\in\mathbb{R}\:\mathrm{with}\:{a}\geqslant\mathrm{8} \\ $$$${x}_{\mathrm{2}} =\frac{\mathrm{1}−\sqrt{{a}+\mathrm{1}}+\sqrt{{a}−\mathrm{2}−\mathrm{2}\sqrt{{a}+\mathrm{1}}}}{\mathrm{2}}\:\in\mathbb{R}\:\mathrm{with}\:{a}\geqslant\mathrm{8} \\ $$$${x}_{\mathrm{3}} =\frac{\mathrm{1}+\sqrt{{a}+\mathrm{1}}−\sqrt{{a}−\mathrm{2}+\mathrm{2}\sqrt{{a}+\mathrm{1}}}}{\mathrm{2}}\:\in\mathbb{R}\:\mathrm{with}\:{a}\geqslant\mathrm{0} \\ $$$${x}_{\mathrm{4}} =\frac{\mathrm{1}+\sqrt{{a}+\mathrm{1}}+\sqrt{{a}−\mathrm{2}+\mathrm{2}\sqrt{{a}+\mathrm{1}}}}{\mathrm{2}}\:\in\mathbb{R}\:\mathrm{with}\:{a}\geqslant\mathrm{0} \\ $$$$ \\ $$$$\mathrm{btw}\:\left(\mathrm{1}\right)\:{x}_{\mathrm{2}} =\frac{\mathrm{1}}{{x}_{\mathrm{1}} }\wedge{x}_{\mathrm{4}} =\frac{\mathrm{1}}{{x}_{\mathrm{3}} }\:\mathrm{because} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{P}_{{a}} \left({x}\right)={x}^{\mathrm{4}} −\mathrm{2}{x}^{\mathrm{3}} −\left({a}−\mathrm{2}\right){x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}\:\mathrm{symmetric} \\ $$$$\:\:\:\:\:\:\:\:\:\:\left(\mathrm{2}\right)\:\mathrm{no}\:\mathrm{real}\:\mathrm{solution}\:\mathrm{for}\:{a}<\mathrm{0}\:\mathrm{which}\:\mathrm{is}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{see}: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{P}_{{a}} \left({x}\right)=\mathrm{0}\:\Rightarrow\:{a}=\frac{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}−\mathrm{1}\right)^{\mathrm{2}} }{{x}^{\mathrm{2}} }\geqslant\mathrm{0} \\ $$
Answered by MJS_new last updated on 30/Aug/21
(x^2 +1)(x−1)^2 −ax^2 =0 x^4 −2x^3 −(a−2)x^2 −2x+1=0 x=t+(1/2) t^4 −((2a−1)/2)t^2 −(a+1)t−((4a−5)/(16))=0 (t^2 −αt−β)(t^2 +αt−γ)=0 matching the coefficients leads to { ((α^2 +β+γ=((2a−1)/2))),((αβ−αγ=a+1)),((βγ=−((4a−5)/(16)))) :} solving (1) and (2) for β and γ ⇒ { ((β=−(α^2 /2)+((a+1)/(2α))+((2a−1)/4))),((γ=−(α^2 /2)−((a+1)/(2α))+((2a−1)/4))),((α^6 −(2a−1)α^4 +(a^2 −1)α^2 −(a+1)^2 =0)) :} α=(√(z+((2a−1)/3))) z^3 −(((a−2)^2 )/3)+((2(a+4)(a^2 −10a−2))/(27))=0 testing factors of ((2(a+4)(a^2 −10a−2))/(27)) ⇒ z=((a+4)/3) ⇒ α=(√(a+1)) ⇒ β=((√(a+1))/2)−(3/4)∧γ=−((√(a+1))/2)−(3/4) ⇒ (t^2 −(√(a+1))t−((2(√(a+1))−3)/4))(t^2 +(√(a+1))t+((2(√(a+1))+3)/4))=0 ⇒ (x^2 −(1+(√(a+1)))x+1)(x^2 −(1−(√(a+1))x+1)=0 ⇒ x=((1+(√(a+1))±(√(a−2+2(√(a+1)))))/2)∨x=((1+(√(a+1))±(√(a−2−2(√(a+1)))))/2) to get 2 real solutions a≥0 to get 4 real solutions a≥8
$$\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}−\mathrm{1}\right)^{\mathrm{2}} −{ax}^{\mathrm{2}} =\mathrm{0} \\ $$$${x}^{\mathrm{4}} −\mathrm{2}{x}^{\mathrm{3}} −\left({a}−\mathrm{2}\right){x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}=\mathrm{0} \\ $$$${x}={t}+\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${t}^{\mathrm{4}} −\frac{\mathrm{2}{a}−\mathrm{1}}{\mathrm{2}}{t}^{\mathrm{2}} −\left({a}+\mathrm{1}\right){t}−\frac{\mathrm{4}{a}−\mathrm{5}}{\mathrm{16}}=\mathrm{0} \\ $$$$\left({t}^{\mathrm{2}} −\alpha{t}−\beta\right)\left({t}^{\mathrm{2}} +\alpha{t}−\gamma\right)=\mathrm{0} \\ $$$$\mathrm{matching}\:\mathrm{the}\:\mathrm{coefficients}\:\mathrm{leads}\:\mathrm{to} \\ $$$$\begin{cases}{\alpha^{\mathrm{2}} +\beta+\gamma=\frac{\mathrm{2}{a}−\mathrm{1}}{\mathrm{2}}}\\{\alpha\beta−\alpha\gamma={a}+\mathrm{1}}\\{\beta\gamma=−\frac{\mathrm{4}{a}−\mathrm{5}}{\mathrm{16}}}\end{cases} \\ $$$$\mathrm{solving}\:\left(\mathrm{1}\right)\:\mathrm{and}\:\left(\mathrm{2}\right)\:\mathrm{for}\:\beta\:\mathrm{and}\:\gamma\:\Rightarrow \\ $$$$\begin{cases}{\beta=−\frac{\alpha^{\mathrm{2}} }{\mathrm{2}}+\frac{{a}+\mathrm{1}}{\mathrm{2}\alpha}+\frac{\mathrm{2}{a}−\mathrm{1}}{\mathrm{4}}}\\{\gamma=−\frac{\alpha^{\mathrm{2}} }{\mathrm{2}}−\frac{{a}+\mathrm{1}}{\mathrm{2}\alpha}+\frac{\mathrm{2}{a}−\mathrm{1}}{\mathrm{4}}}\\{\alpha^{\mathrm{6}} −\left(\mathrm{2}{a}−\mathrm{1}\right)\alpha^{\mathrm{4}} +\left({a}^{\mathrm{2}} −\mathrm{1}\right)\alpha^{\mathrm{2}} −\left({a}+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{0}}\end{cases} \\ $$$$\alpha=\sqrt{{z}+\frac{\mathrm{2}{a}−\mathrm{1}}{\mathrm{3}}} \\ $$$${z}^{\mathrm{3}} −\frac{\left({a}−\mathrm{2}\right)^{\mathrm{2}} }{\mathrm{3}}+\frac{\mathrm{2}\left({a}+\mathrm{4}\right)\left({a}^{\mathrm{2}} −\mathrm{10}{a}−\mathrm{2}\right)}{\mathrm{27}}=\mathrm{0} \\ $$$$\mathrm{testing}\:\mathrm{factors}\:\mathrm{of}\:\frac{\mathrm{2}\left({a}+\mathrm{4}\right)\left({a}^{\mathrm{2}} −\mathrm{10}{a}−\mathrm{2}\right)}{\mathrm{27}}\:\Rightarrow \\ $$$${z}=\frac{{a}+\mathrm{4}}{\mathrm{3}}\:\Rightarrow \\ $$$$\alpha=\sqrt{{a}+\mathrm{1}}\:\Rightarrow \\ $$$$\beta=\frac{\sqrt{{a}+\mathrm{1}}}{\mathrm{2}}−\frac{\mathrm{3}}{\mathrm{4}}\wedge\gamma=−\frac{\sqrt{{a}+\mathrm{1}}}{\mathrm{2}}−\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\Rightarrow \\ $$$$\left({t}^{\mathrm{2}} −\sqrt{{a}+\mathrm{1}}{t}−\frac{\mathrm{2}\sqrt{{a}+\mathrm{1}}−\mathrm{3}}{\mathrm{4}}\right)\left({t}^{\mathrm{2}} +\sqrt{{a}+\mathrm{1}}{t}+\frac{\mathrm{2}\sqrt{{a}+\mathrm{1}}+\mathrm{3}}{\mathrm{4}}\right)=\mathrm{0} \\ $$$$\Rightarrow \\ $$$$\left({x}^{\mathrm{2}} −\left(\mathrm{1}+\sqrt{{a}+\mathrm{1}}\right){x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} −\left(\mathrm{1}−\sqrt{{a}+\mathrm{1}}{x}+\mathrm{1}\right)=\mathrm{0}\right. \\ $$$$\Rightarrow \\ $$$${x}=\frac{\mathrm{1}+\sqrt{{a}+\mathrm{1}}\pm\sqrt{{a}−\mathrm{2}+\mathrm{2}\sqrt{{a}+\mathrm{1}}}}{\mathrm{2}}\vee{x}=\frac{\mathrm{1}+\sqrt{{a}+\mathrm{1}}\pm\sqrt{{a}−\mathrm{2}−\mathrm{2}\sqrt{{a}+\mathrm{1}}}}{\mathrm{2}} \\ $$$$\mathrm{to}\:\mathrm{get}\:\mathrm{2}\:\mathrm{real}\:\mathrm{solutions}\:{a}\geqslant\mathrm{0} \\ $$$$\mathrm{to}\:\mathrm{get}\:\mathrm{4}\:\mathrm{real}\:\mathrm{solutions}\:{a}\geqslant\mathrm{8} \\ $$
Answered by EDWIN88 last updated on 01/Sep/21
we have (x^2 +1)(x^2 −2x+1)−ax^2 =0 Dividing by x^2 yields (x+(1/x))(x−2+(1/x))−a=0 let x+(1/x)=h ⇒h^2 −2h−a=0 it follows that h=x+(1/x)=((2±(√(4+4a)))/2)=1±(√a) which in turn implies that if a≥0 then the polynomial P_a (x) has real zeros x_(1,2) =((1+(√(1+a)) ±(√(a+2(√(1+a)) −2)))/2) in addition if a≥8 then P_a (x) also has the real zeros x_(3,4) =((1−(√(1+a)) ± (√(a−2(√(1+a))−2)))/2)
$${we}\:{have}\:\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}\right)−{ax}^{\mathrm{2}} =\mathrm{0} \\ $$$${Dividing}\:{by}\:{x}^{\mathrm{2}} \:{yields}\: \\ $$$$\:\left({x}+\frac{\mathrm{1}}{{x}}\right)\left({x}−\mathrm{2}+\frac{\mathrm{1}}{{x}}\right)−{a}=\mathrm{0} \\ $$$$\:{let}\:{x}+\frac{\mathrm{1}}{{x}}={h}\:\Rightarrow{h}^{\mathrm{2}} −\mathrm{2}{h}−{a}=\mathrm{0} \\ $$$$\:{it}\:{follows}\:{that}\:{h}={x}+\frac{\mathrm{1}}{{x}}=\frac{\mathrm{2}\pm\sqrt{\mathrm{4}+\mathrm{4}{a}}}{\mathrm{2}}=\mathrm{1}\pm\sqrt{{a}} \\ $$$$\:{which}\:{in}\:{turn}\:{implies}\:{that}\:{if}\: \\ $$$$\:{a}\geqslant\mathrm{0}\:{then}\:{the}\:{polynomial}\: \\ $$$$\:{P}_{{a}} \left({x}\right)\:{has}\:{real}\:{zeros}\: \\ $$$$\:{x}_{\mathrm{1},\mathrm{2}} =\frac{\mathrm{1}+\sqrt{\mathrm{1}+{a}}\:\pm\sqrt{{a}+\mathrm{2}\sqrt{\mathrm{1}+{a}}\:−\mathrm{2}}}{\mathrm{2}} \\ $$$${in}\:{addition}\:{if}\:{a}\geqslant\mathrm{8}\:{then}\:{P}_{{a}} \left({x}\right)\:{also} \\ $$$$\:{has}\:{the}\:{real}\:{zeros}\: \\ $$$$\:{x}_{\mathrm{3},\mathrm{4}} =\frac{\mathrm{1}−\sqrt{\mathrm{1}+{a}}\:\pm\:\sqrt{{a}−\mathrm{2}\sqrt{\mathrm{1}+{a}}−\mathrm{2}}}{\mathrm{2}} \\ $$

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