Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 90793 by kelum last updated on 26/Apr/20

a+b+c+d=4 a^2 +b^2 +c^2 +d^2 =10 a^3 +b^3 +c^3 +d^3 =22 a^4 +b^4 +c^4 +d^4 = ?

$${a}+{b}+{c}+{d}=\mathrm{4} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{d}^{\mathrm{2}} =\mathrm{10} \\ $$$${a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} +{d}^{\mathrm{3}} =\mathrm{22} \\ $$$${a}^{\mathrm{4}} +{b}^{\mathrm{4}} +{c}^{\mathrm{4}} +{d}^{\mathrm{4}} =\:? \\ $$

Commented by mr W last updated on 26/Apr/20

you are sure the question is correct? we have four unknowns but only three eqn., so we can not get an unique solution.

$${you}\:{are}\:{sure}\:{the}\:{question}\:{is}\:{correct}? \\ $$$${we}\:{have}\:{four}\:{unknowns}\:{but}\:{only} \\ $$$${three}\:{eqn}.,\:{so}\:{we}\:{can}\:{not}\:{get}\:{an}\:{unique} \\ $$$${solution}. \\ $$

Commented by kelum last updated on 26/Apr/20

Yes its correct. I know the answers Just dont know how to get tbe answers.

$${Yes}\:{its}\:{correct}.\:{I}\:{know}\:{the}\:{answers} \\ $$$${Just}\:{dont}\:{know}\:{how}\:{to}\:{get}\:{tbe}\: \\ $$$${answers}. \\ $$$$ \\ $$

Commented by mr W last updated on 26/Apr/20

what is the answer? i got no unique solution. we can only say a^4 +b^4 +c^4 +d^4 ≥49

$${what}\:{is}\:{the}\:{answer}? \\ $$$${i}\:{got}\:{no}\:{unique}\:{solution}. \\ $$$${we}\:{can}\:{only}\:{say} \\ $$$${a}^{\mathrm{4}} +{b}^{\mathrm{4}} +{c}^{\mathrm{4}} +{d}^{\mathrm{4}} \geqslant\mathrm{49} \\ $$

Commented by mr W last updated on 29/Apr/20

if you know the answer, please share it!

$${if}\:{you}\:{know}\:{the}\:{answer},\:{please}\:{share} \\ $$$${it}! \\ $$

Answered by mr W last updated on 26/Apr/20

p_1 =a+b+c=4−d p_2 =a^2 +b^2 +c^2 =10−d^2 p_3 =a^3 +b^3 +c^3 =22−d^3 ⇒e_1 =4−d p_2 =e_1 ^2 −2e_2 ⇒10−d^2 =(4−d)^2 −2e_2 ⇒e_2 =d^2 −4d+3 p_3 =e_1 ^3 −3e_1 e_2 +3e_3 ⇒22−d^3 =(4−d)^3 −3(4−d)(d^2 −4d+3)+3e_3 ⇒22−d^3 =(4−d)(−2d^2 +4d+7)+3e_3 ⇒e_3 =−d^3 +4d^2 −3d−2 p_4 =e_1 ^4 −4e_1 ^2 e_2 +4e_1 e_3 +2e_2 ^2 −4e_4 p_4 =(4−d)^4 −4(4−d)^2 (d^2 −4d+3)+4(4−d)(−d^3 +4d^2 −3d−2)+2(d^2 −4d+3)^2 p_4 =3d^4 −16d^3 +12d^2 +8d+50 a^4 +b^4 +c^4 =3d^4 −16d^3 +12d^2 +8d+50 a^4 +b^4 +c^4 +d^4 =4d(d−2)(d^2 −2d−1)+50≠constant

$${p}_{\mathrm{1}} ={a}+{b}+{c}=\mathrm{4}−{d} \\ $$$${p}_{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} =\mathrm{10}−{d}^{\mathrm{2}} \\ $$$${p}_{\mathrm{3}} ={a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} =\mathrm{22}−{d}^{\mathrm{3}} \\ $$$$\Rightarrow{e}_{\mathrm{1}} =\mathrm{4}−{d} \\ $$$${p}_{\mathrm{2}} ={e}_{\mathrm{1}} ^{\mathrm{2}} −\mathrm{2}{e}_{\mathrm{2}} \: \\ $$$$\:\Rightarrow\mathrm{10}−{d}^{\mathrm{2}} =\left(\mathrm{4}−{d}\right)^{\mathrm{2}} −\mathrm{2}{e}_{\mathrm{2}} \\ $$$$\:\Rightarrow{e}_{\mathrm{2}} ={d}^{\mathrm{2}} −\mathrm{4}{d}+\mathrm{3} \\ $$$${p}_{\mathrm{3}} ={e}_{\mathrm{1}} ^{\mathrm{3}} −\mathrm{3}{e}_{\mathrm{1}} {e}_{\mathrm{2}} +\mathrm{3}{e}_{\mathrm{3}} \\ $$$$\Rightarrow\mathrm{22}−{d}^{\mathrm{3}} =\left(\mathrm{4}−{d}\right)^{\mathrm{3}} −\mathrm{3}\left(\mathrm{4}−{d}\right)\left({d}^{\mathrm{2}} −\mathrm{4}{d}+\mathrm{3}\right)+\mathrm{3}{e}_{\mathrm{3}} \\ $$$$\Rightarrow\mathrm{22}−{d}^{\mathrm{3}} =\left(\mathrm{4}−{d}\right)\left(−\mathrm{2}{d}^{\mathrm{2}} +\mathrm{4}{d}+\mathrm{7}\right)+\mathrm{3}{e}_{\mathrm{3}} \\ $$$$\Rightarrow{e}_{\mathrm{3}} =−{d}^{\mathrm{3}} +\mathrm{4}{d}^{\mathrm{2}} −\mathrm{3}{d}−\mathrm{2} \\ $$$${p}_{\mathrm{4}} ={e}_{\mathrm{1}} ^{\mathrm{4}} −\mathrm{4}{e}_{\mathrm{1}} ^{\mathrm{2}} {e}_{\mathrm{2}} +\mathrm{4}{e}_{\mathrm{1}} {e}_{\mathrm{3}} +\mathrm{2}{e}_{\mathrm{2}} ^{\mathrm{2}} −\mathrm{4}{e}_{\mathrm{4}} \\ $$$${p}_{\mathrm{4}} =\left(\mathrm{4}−{d}\right)^{\mathrm{4}} −\mathrm{4}\left(\mathrm{4}−{d}\right)^{\mathrm{2}} \left({d}^{\mathrm{2}} −\mathrm{4}{d}+\mathrm{3}\right)+\mathrm{4}\left(\mathrm{4}−{d}\right)\left(−{d}^{\mathrm{3}} +\mathrm{4}{d}^{\mathrm{2}} −\mathrm{3}{d}−\mathrm{2}\right)+\mathrm{2}\left({d}^{\mathrm{2}} −\mathrm{4}{d}+\mathrm{3}\right)^{\mathrm{2}} \\ $$$${p}_{\mathrm{4}} =\mathrm{3}{d}^{\mathrm{4}} −\mathrm{16}{d}^{\mathrm{3}} +\mathrm{12}{d}^{\mathrm{2}} +\mathrm{8}{d}+\mathrm{50} \\ $$$${a}^{\mathrm{4}} +{b}^{\mathrm{4}} +{c}^{\mathrm{4}} =\mathrm{3}{d}^{\mathrm{4}} −\mathrm{16}{d}^{\mathrm{3}} +\mathrm{12}{d}^{\mathrm{2}} +\mathrm{8}{d}+\mathrm{50} \\ $$$${a}^{\mathrm{4}} +{b}^{\mathrm{4}} +{c}^{\mathrm{4}} +{d}^{\mathrm{4}} =\mathrm{4}{d}\left({d}−\mathrm{2}\right)\left({d}^{\mathrm{2}} −\mathrm{2}{d}−\mathrm{1}\right)+\mathrm{50}\neq{constant} \\ $$

Commented by mr W last updated on 26/Apr/20

or p_1 =a+b+c+d=4 p_2 =a^2 +b^2 +c^2 +d^2 =10 p_3 =a^3 +b^3 +c^3 +d^3 =22 p_1 =e_1 =4 p_2 =e_1 ^2 −2e_2 ⇒10=4^2 −2e_2 ⇒e_2 =3 p_3 =e_1 ^3 −3e_1 e_2 +3e_3 ⇒22=64−36+3e_3 ⇒e_3 =−2 p_4 =e_1 ^4 −4e_1 ^2 e_2 +4e_1 e_3 +2e_2 ^2 −4e_4 =4^4 −4×4^2 ×3+4×4×(−2)+2×3^2 −4e_4 =50−4abcd ⇒a^4 +b^4 +c^4 +d^4 =50−4abcd≠constant

$${or} \\ $$$${p}_{\mathrm{1}} ={a}+{b}+{c}+{d}=\mathrm{4} \\ $$$${p}_{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{d}^{\mathrm{2}} =\mathrm{10} \\ $$$${p}_{\mathrm{3}} ={a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} +{d}^{\mathrm{3}} =\mathrm{22} \\ $$$${p}_{\mathrm{1}} ={e}_{\mathrm{1}} =\mathrm{4} \\ $$$${p}_{\mathrm{2}} ={e}_{\mathrm{1}} ^{\mathrm{2}} −\mathrm{2}{e}_{\mathrm{2}} \:\Rightarrow\mathrm{10}=\mathrm{4}^{\mathrm{2}} −\mathrm{2}{e}_{\mathrm{2}} \:\Rightarrow{e}_{\mathrm{2}} =\mathrm{3} \\ $$$${p}_{\mathrm{3}} ={e}_{\mathrm{1}} ^{\mathrm{3}} −\mathrm{3}{e}_{\mathrm{1}} {e}_{\mathrm{2}} +\mathrm{3}{e}_{\mathrm{3}} \:\Rightarrow\mathrm{22}=\mathrm{64}−\mathrm{36}+\mathrm{3}{e}_{\mathrm{3}} \:\Rightarrow{e}_{\mathrm{3}} =−\mathrm{2} \\ $$$${p}_{\mathrm{4}} ={e}_{\mathrm{1}} ^{\mathrm{4}} −\mathrm{4}{e}_{\mathrm{1}} ^{\mathrm{2}} {e}_{\mathrm{2}} +\mathrm{4}{e}_{\mathrm{1}} {e}_{\mathrm{3}} +\mathrm{2}{e}_{\mathrm{2}} ^{\mathrm{2}} −\mathrm{4}{e}_{\mathrm{4}} \\ $$$$=\mathrm{4}^{\mathrm{4}} −\mathrm{4}×\mathrm{4}^{\mathrm{2}} ×\mathrm{3}+\mathrm{4}×\mathrm{4}×\left(−\mathrm{2}\right)+\mathrm{2}×\mathrm{3}^{\mathrm{2}} −\mathrm{4}{e}_{\mathrm{4}} \\ $$$$=\mathrm{50}−\mathrm{4}{abcd} \\ $$$$\Rightarrow{a}^{\mathrm{4}} +{b}^{\mathrm{4}} +{c}^{\mathrm{4}} +{d}^{\mathrm{4}} =\mathrm{50}−\mathrm{4}{abcd}\neq{constant} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com