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Question-7299

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Question Number 7299 by Tawakalitu. last updated on 22/Aug/16
Commented by sou1618 last updated on 22/Aug/16
f(x)=x^4 +2x^3 −x^2 +2x+1 x=y−(2/4) f(y−(1/2))=(y−1/2)^4 +2(y−1/2)^3 −(y−1/2)^2 +2(y−1/2)+1 f(y−1/2)=(y^4 −2y^3 +(3/2)y^2 −(1/2)y+(1/(16)))+(2y^3 −3y^2 +(3/2)y−(1/4)) +(−y^2 +y−(1/4))+(2y−1)+1 f(y−1/2)=y^4 −(5/2)y^2 +4y−(7/(16)) (a,b,c∈Z) (4y^2 +a)^2 −16f(y−1/2)=(by+c)^2 y^2 (8a+40)−64y+a^2 +7=b^2 y^2 +2bcy+c^2 { ((b^2 =8a+40)),((2bc=−64)),((c^2 =a^2 +7)) :} (1)c^2 −a^2 =7 (c−a)(c+a)=1×7,(−1)×(−7) (c,a)=(4,3),(−4,−3) (2)bc=−32 if c=4...b=−8 (a,b,c)=(3,−8,4) if c=−4...b=8 (a,b,c)=(−3,8,−4) (3)b^2 =8a+40 b=±8 64=8a+40⇒a=3 so (a,b,c)=(3,−8,4) (4y^2 +3)^2 −16f(y−1/2)=(−8y+4)^2 when f(y−1/2)=0 (4y^2 +3)^2 =(−8y+4)^2 4y^2 +3=±(−8y+4) { ((4y^2 +8y−1=0)),((4y^2 −8y+7=0)) :} { ((y=((−4±(√(16+4)))/4))),((y=((4±(√(16−28)))/4))) :} y=((−2±(√5))/2),((2±(√(−3)))/2) ⇒ x=((−2±(√5))/2)−(1/2) , ((2±(√(−3)))/2)−(1/2) x=((−3±(√5))/2),((1±(√3)i)/2)
$${f}\left({x}\right)={x}^{\mathrm{4}} +\mathrm{2}{x}^{\mathrm{3}} −{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1} \\ $$$${x}={y}−\frac{\mathrm{2}}{\mathrm{4}} \\ $$$${f}\left({y}−\frac{\mathrm{1}}{\mathrm{2}}\right)=\left({y}−\mathrm{1}/\mathrm{2}\right)^{\mathrm{4}} +\mathrm{2}\left({y}−\mathrm{1}/\mathrm{2}\right)^{\mathrm{3}} −\left({y}−\mathrm{1}/\mathrm{2}\right)^{\mathrm{2}} +\mathrm{2}\left({y}−\mathrm{1}/\mathrm{2}\right)+\mathrm{1} \\ $$$${f}\left({y}−\mathrm{1}/\mathrm{2}\right)=\left({y}^{\mathrm{4}} −\mathrm{2}{y}^{\mathrm{3}} +\frac{\mathrm{3}}{\mathrm{2}}{y}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}{y}+\frac{\mathrm{1}}{\mathrm{16}}\right)+\left(\mathrm{2}{y}^{\mathrm{3}} −\mathrm{3}{y}^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{2}}{y}−\frac{\mathrm{1}}{\mathrm{4}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\left(−{y}^{\mathrm{2}} +{y}−\frac{\mathrm{1}}{\mathrm{4}}\right)+\left(\mathrm{2}{y}−\mathrm{1}\right)+\mathrm{1} \\ $$$${f}\left({y}−\mathrm{1}/\mathrm{2}\right)={y}^{\mathrm{4}} −\frac{\mathrm{5}}{\mathrm{2}}{y}^{\mathrm{2}} +\mathrm{4}{y}−\frac{\mathrm{7}}{\mathrm{16}} \\ $$$$ \\ $$$$\left({a},{b},{c}\in\mathbb{Z}\right) \\ $$$$\left(\mathrm{4}{y}^{\mathrm{2}} +{a}\right)^{\mathrm{2}} −\mathrm{16}{f}\left({y}−\mathrm{1}/\mathrm{2}\right)=\left({by}+{c}\right)^{\mathrm{2}} \\ $$$${y}^{\mathrm{2}} \left(\mathrm{8}{a}+\mathrm{40}\right)−\mathrm{64}{y}+{a}^{\mathrm{2}} +\mathrm{7}={b}^{\mathrm{2}} {y}^{\mathrm{2}} +\mathrm{2}{bcy}+{c}^{\mathrm{2}} \\ $$$$\begin{cases}{{b}^{\mathrm{2}} =\mathrm{8}{a}+\mathrm{40}}\\{\mathrm{2}{bc}=−\mathrm{64}}\\{{c}^{\mathrm{2}} ={a}^{\mathrm{2}} +\mathrm{7}}\end{cases} \\ $$$$\left(\mathrm{1}\right){c}^{\mathrm{2}} −{a}^{\mathrm{2}} =\mathrm{7} \\ $$$$\:\:\:\left({c}−{a}\right)\left({c}+{a}\right)=\mathrm{1}×\mathrm{7},\left(−\mathrm{1}\right)×\left(−\mathrm{7}\right) \\ $$$$\:\:\:\:\left({c},{a}\right)=\left(\mathrm{4},\mathrm{3}\right),\left(−\mathrm{4},−\mathrm{3}\right) \\ $$$$\left(\mathrm{2}\right){bc}=−\mathrm{32} \\ $$$$\:\:{if}\:{c}=\mathrm{4}…{b}=−\mathrm{8} \\ $$$$\:\:\:\:\:\:\left({a},{b},{c}\right)=\left(\mathrm{3},−\mathrm{8},\mathrm{4}\right) \\ $$$$\:\:{if}\:{c}=−\mathrm{4}…{b}=\mathrm{8} \\ $$$$\:\:\:\:\:\:\left({a},{b},{c}\right)=\left(−\mathrm{3},\mathrm{8},−\mathrm{4}\right) \\ $$$$\left(\mathrm{3}\right){b}^{\mathrm{2}} =\mathrm{8}{a}+\mathrm{40} \\ $$$$\:\:{b}=\pm\mathrm{8} \\ $$$$\:\:\mathrm{64}=\mathrm{8}{a}+\mathrm{40}\Rightarrow{a}=\mathrm{3} \\ $$$${so} \\ $$$$\left({a},{b},{c}\right)=\left(\mathrm{3},−\mathrm{8},\mathrm{4}\right) \\ $$$$ \\ $$$$\left(\mathrm{4}{y}^{\mathrm{2}} +\mathrm{3}\right)^{\mathrm{2}} −\mathrm{16}{f}\left({y}−\mathrm{1}/\mathrm{2}\right)=\left(−\mathrm{8}{y}+\mathrm{4}\right)^{\mathrm{2}} \\ $$$${when}\:{f}\left({y}−\mathrm{1}/\mathrm{2}\right)=\mathrm{0} \\ $$$$\left(\mathrm{4}{y}^{\mathrm{2}} +\mathrm{3}\right)^{\mathrm{2}} =\left(−\mathrm{8}{y}+\mathrm{4}\right)^{\mathrm{2}} \\ $$$$\mathrm{4}{y}^{\mathrm{2}} +\mathrm{3}=\pm\left(−\mathrm{8}{y}+\mathrm{4}\right) \\ $$$$\begin{cases}{\mathrm{4}{y}^{\mathrm{2}} +\mathrm{8}{y}−\mathrm{1}=\mathrm{0}}\\{\mathrm{4}{y}^{\mathrm{2}} −\mathrm{8}{y}+\mathrm{7}=\mathrm{0}}\end{cases} \\ $$$$\begin{cases}{{y}=\frac{−\mathrm{4}\pm\sqrt{\mathrm{16}+\mathrm{4}}}{\mathrm{4}}}\\{{y}=\frac{\mathrm{4}\pm\sqrt{\mathrm{16}−\mathrm{28}}}{\mathrm{4}}}\end{cases} \\ $$$${y}=\frac{−\mathrm{2}\pm\sqrt{\mathrm{5}}}{\mathrm{2}},\frac{\mathrm{2}\pm\sqrt{−\mathrm{3}}}{\mathrm{2}} \\ $$$$\Rightarrow \\ $$$${x}=\frac{−\mathrm{2}\pm\sqrt{\mathrm{5}}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\:,\:\frac{\mathrm{2}\pm\sqrt{−\mathrm{3}}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${x}=\frac{−\mathrm{3}\pm\sqrt{\mathrm{5}}}{\mathrm{2}},\frac{\mathrm{1}\pm\sqrt{\mathrm{3}}{i}}{\mathrm{2}} \\ $$$$ \\ $$
Commented by Tawakalitu. last updated on 22/Aug/16
Great job. thank you sir for your help.
$${Great}\:{job}.\:{thank}\:{you}\:{sir}\:{for}\:{your}\:{help}. \\ $$
Commented by Yozzia last updated on 22/Aug/16
Nice!
$$ \\ $$$${Nice}! \\ $$
Commented by sandy_suhendra last updated on 22/Aug/16
because : (x^2 +x+1)^2 = x^4 +2x^3 +3x^2 +2x+1 so : x^4 +2x^3 −x^2 +2x+1=0 (x^2 +x+1)^2 − 4x^2 = 0 (x^2 +x+1+2x)(x^2 +x+1−2x)=0 (x^2 +3x+1)(x^2 −x+1)=0 x^2 +3x+1=0 ⇒ x = ((−3±(√5))/2) x^2 −x+1=0 ⇒ x = ((1±i(√3))/2)
$${because}\::\:\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)^{\mathrm{2}} \:=\:{x}^{\mathrm{4}} +\mathrm{2}{x}^{\mathrm{3}} +\mathrm{3}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1} \\ $$$${so}\::\:{x}^{\mathrm{4}} +\mathrm{2}{x}^{\mathrm{3}} −{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)^{\mathrm{2}} −\:\mathrm{4}{x}^{\mathrm{2}} \:=\:\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\left({x}^{\mathrm{2}} +{x}+\mathrm{1}+\mathrm{2}{x}\right)\left({x}^{\mathrm{2}} +{x}+\mathrm{1}−\mathrm{2}{x}\right)=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\left({x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)=\mathrm{0} \\ $$$${x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{1}=\mathrm{0}\:\Rightarrow\:{x}\:=\:\frac{−\mathrm{3}\pm\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} −{x}+\mathrm{1}=\mathrm{0}\:\Rightarrow\:{x}\:=\:\frac{\mathrm{1}\pm{i}\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$
Commented by sou1618 last updated on 23/Aug/16
wow nice!!
$${wow}\:{nice}!! \\ $$
Commented by Tawakalitu. last updated on 23/Aug/16
Interesting. Thanks so much.
$${Interesting}.\:{Thanks}\:{so}\:{much}. \\ $$

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