Math Question and Answers


Question and Answers Forum

All Questions Topic List
None Questions
Previous in All Question Next in All Question
Previous in None Next in None
Question Number 70847 by A8;15: last updated on 08/Oct/19

	Answered by mind is power last updated on 08/Oct/19
	${ ((x^2 +y^2 =2)),((2x^4 +xy^3 +x^2 y^2 −x^3 y=3)) :}∙EQ1 we have x≠0 EQ1⇔ { ((1+((y/x))^2 =(2/x^2 ))),((2+((y/x))^3 +((y/x))^2 −(y/x)=(3/x^4 ))) :}_ let u=(y/x) and z=(1/x^2 )⇒ { ((y^2 =(u^2 /z^2 ))),((x^2 =(1/z))) :} EQ1⇔ { ((1+u^2 =2z)),((2+u^3 +u^2 −u=3z^2 )) :} ⇒2+u^3 +u^2 −u=(3/4)(1+u^2 )^2 ⇒8+4u^3 +4u^2 −4u=3u^4 +6u^2 +3 ⇒3u^4 −4u^3 +2u^2 +4u−5=0 u=1 is solution ⇒(u−1)(3u^3 −u^2 +u+5)=0 3u^3 −u^2 +u+5=0 we see u=−1 is solution ⇒3u^3 −u^2 +u+5=(u+1)(3u^2 −4u+5) 3u^2 −4u+5=0 ⇒16−60=−48 u_3 =((4−4i(√3))/6) u_4 =((4+4i(√3))/6) for u=1or −1 ,z=((1+u^2 )/2)=((1+1)/2)=1 x^2 =(1/z)=1⇒x∈{−1,1} y^2 =(1/1)⇒y∈{1,−1} S′={(1,1),(1,−1),(−1,1),(−1,−1)} be contiued withe U_3 and u_4$
	${\begin{cases} x^{2} + y^{2} = 2 \\ 2 x^{4} + {x y}^{3} + x^{2} y^{2} - x^{3} y = 3 \end{cases} \cdot E Q 1$ $w e h a v e x \neq 0$ $E Q 1 \Leftrightarrow {\begin{cases} 1 + {(\frac{y}{x})}^{2} = \frac{2}{x^{2}} \\ 2 + {(\frac{y}{x})}^{3} + {(\frac{y}{x})}^{2} - \frac{y}{x} = \frac{3}{x^{4}} \end{cases}_{}$ $l e t u = \frac{y}{x} a n d z = \frac{1}{x^{2}} \Rightarrow {\begin{cases} y^{2} = \frac{u^{2}}{z^{2}} \\ x^{2} = \frac{1}{z} \end{cases}$ $E Q 1 \Leftrightarrow {\begin{cases} 1 + u^{2} = 2 z \\ 2 + u^{3} + u^{2} - u = 3 z^{2} \end{cases}$ $\Rightarrow 2 + u^{3} + u^{2} - u = \frac{3}{4} {(1 + u^{2})}^{2}$ $\Rightarrow 8 + 4 u^{3} + 4 u^{2} - 4 u = 3 u^{4} + 6 u^{2} + 3$ $\Rightarrow 3 u^{4} - 4 u^{3} + 2 u^{2} + 4 u - 5 = 0$ $u = 1 i s s o l u t i o n$ $\Rightarrow (u - 1) (3 u^{3} - u^{2} + u + 5) = 0$ $3 u^{3} - u^{2} + u + 5 = 0$ $w e s e e u = - 1 i s s o l u t i o n$ $\Rightarrow 3 u^{3} - u^{2} + u + 5 = (u + 1) (3 u^{2} - 4 u + 5)$ $3 u^{2} - 4 u + 5 = 0$ $\Rightarrow 16 - 60 = - 48$ $u_{3} = \frac{4 - 4 i \sqrt{3}}{6}$ $u_{4} = \frac{4 + 4 i \sqrt{3}}{6}$ $f o r u = 1 o r - 1, z = \frac{1 + u^{2}}{2} = \frac{1 + 1}{2} = 1$ $x^{2} = \frac{1}{z} = 1 \Rightarrow x \in {- 1, 1}$ $y^{2} = \frac{1}{1} \Rightarrow y \in {1, - 1}$ $S^{'} = {(1, 1), (1, - 1), (- 1, 1), (- 1, - 1)}$ $b e c o n t i u e d w i t h e U_{3} a n d u_{4}$
	Commented by MJS last updated on 08/Oct/19

	${you}^{'} re right but {there}^{'} s an easier path$
	Commented by mind is power last updated on 08/Oct/19

	$y e a h, i s e e y o u r s o l u t i o n S i r i t s b e t t e r$
	Commented by MJS last updated on 08/Oct/19

	$thank you, but what I like in here is the$ $variety of ideas to solve each problem$
	Answered by MJS last updated on 08/Oct/19

	$x^{2} + y^{2} = 2$ $2 x^{4} - x^{3} y + x^{2} y^{2} + {x y}^{3} = 3$ $let y = p x$ $(p^{2} + 1) x^{2} = 2 \Rightarrow x^{2} = \frac{2}{p^{2} + 1}$ $(p + 2) (p^{2} - p + 1) x^{4} = 3 \Rightarrow$ $\Rightarrow p^{4} - \frac{4}{3} p^{3} + \frac{2}{3} p^{2} + \frac{4}{3} p - \frac{5}{3} = 0$ $(p - 1) (p + 1) (p^{2} - \frac{4}{3} p + \frac{5}{3}) = 0$ $\Rightarrow p = \pm 1 \lor p = \frac{2}{3} \pm \frac{\sqrt{11}}{3} i$ $\Rightarrow x = \pm 1; y = \pm 1$ $and we can also calculate the complex solutions$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum