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 142232 by liberty last updated on 28/May/21

Answered by MJS_new last updated on 28/May/21

1. x=0 ⇒ y^3 =1∧y^4 −y=0 ⇒ y=1∨y=−(1/2)±((√3)/2)i 2. y=0 ... ⇒ x=1∨x=−(1/2)±((√3)/2)i 3. y=x ... ⇒ x=y=1∨x=y=−(1/2)±((√3)/2)i 4. y=px∧x≠0∧p≠0∧p≠1 (p^3 −p^2 +1)x^3 =1 ⇒ x^3 =(1/(p^3 −p^2 +1)) (4−p^4 )x^4 =(4−p)x ⇒ x^3 =((4−p)/(4−p^4 )) p^3 −p^2 +1=((4−p^4 )/(4−p)) 5p^3 −4p^2 −p=0 p=−(1/5) ⇒ x=(5/( ((119))^(1/3) ))∧y=−(1/( ((119))^(1/3) ))∨x=(5/( ((119))^(1/3) ))(−(1/2)±((√3)/2)i)∧y=−(1/( ((119))^(1/3) ))(−(1/2)±((√3)/2)i)

$$\mathrm{1}.\:{x}=\mathrm{0} \\ $$$$\Rightarrow\:{y}^{\mathrm{3}} =\mathrm{1}\wedge{y}^{\mathrm{4}} −{y}=\mathrm{0}\:\Rightarrow\:{y}=\mathrm{1}\vee{y}=−\frac{\mathrm{1}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i} \\ $$$$\mathrm{2}.\:{y}=\mathrm{0}\:...\:\Rightarrow\:{x}=\mathrm{1}\vee{x}=−\frac{\mathrm{1}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i} \\ $$$$\mathrm{3}.\:{y}={x}\:...\:\Rightarrow\:{x}={y}=\mathrm{1}\vee{x}={y}=−\frac{\mathrm{1}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i} \\ $$$$\mathrm{4}.\:{y}={px}\wedge{x}\neq\mathrm{0}\wedge{p}\neq\mathrm{0}\wedge{p}\neq\mathrm{1} \\ $$$$\left({p}^{\mathrm{3}} −{p}^{\mathrm{2}} +\mathrm{1}\right){x}^{\mathrm{3}} =\mathrm{1}\:\Rightarrow\:{x}^{\mathrm{3}} =\frac{\mathrm{1}}{{p}^{\mathrm{3}} −{p}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\left(\mathrm{4}−{p}^{\mathrm{4}} \right){x}^{\mathrm{4}} =\left(\mathrm{4}−{p}\right){x}\:\Rightarrow\:{x}^{\mathrm{3}} =\frac{\mathrm{4}−{p}}{\mathrm{4}−{p}^{\mathrm{4}} } \\ $$$${p}^{\mathrm{3}} −{p}^{\mathrm{2}} +\mathrm{1}=\frac{\mathrm{4}−{p}^{\mathrm{4}} }{\mathrm{4}−{p}} \\ $$$$\mathrm{5}{p}^{\mathrm{3}} −\mathrm{4}{p}^{\mathrm{2}} −{p}=\mathrm{0} \\ $$$${p}=−\frac{\mathrm{1}}{\mathrm{5}}\:\Rightarrow\:{x}=\frac{\mathrm{5}}{\:\sqrt[{\mathrm{3}}]{\mathrm{119}}}\wedge{y}=−\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\mathrm{119}}}\vee{x}=\frac{\mathrm{5}}{\:\sqrt[{\mathrm{3}}]{\mathrm{119}}}\left(−\frac{\mathrm{1}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}\right)\wedge{y}=−\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\mathrm{119}}}\left(−\frac{\mathrm{1}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}\right) \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com