Math Question and Answers


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 192720 by pascal889 last updated on 25/May/23

	Answered by Frix last updated on 25/May/23

	$x = p - q \land y = p + q$ $\Rightarrow$ $5 p^{4} + 6 p^{2} q^{2} + 5 q^{4} = 109$ $3 p^{2} + q^{2} = 13 \Rightarrow q^{2} = 13 - 3 p^{2}$ $Inserting in 1^{st} equation$ $p^{4} - \frac{39 p^{2}}{4} + 23 = 0$ $p = \pm 2 \lor p = \pm \frac{\sqrt{23}}{2}$ $The rest is easy$
	Answered by BaliramKumar last updated on 25/May/23

	$x^{4} + 3 x^{2} y^{2} + y^{4} = 109 . . . . . . . (i)$ $x^{2} + y^{2} + x y = 13 . . . . . . (i i)$ $x^{4} + 3 x^{2} y^{2} + y^{4} = 109$ ${(x^{2} + y^{2})}^{2} - 2 x^{2} y^{2} + 3 x^{2} y^{2} = 109$ ${(x^{2} + y^{2})}^{2} + x^{2} y^{2} = 109$ ${(x^{2} + y^{2} + x y)}^{2} - 2 x y (x^{2} + y^{2}) = 109$ ${(13)}^{2} - 2 x y (x^{2} + y^{2}) = 109$ $169 - 2 x y (x^{2} + y^{2}) = 109$ $x y (x^{2} + y^{2}) = 30 . . . . . . (i i i)$ $p u t x^{2} + y^{2} = α x y = β$ $α + β = 13 α β = 30$ $t^{2} - 13 t + 30 = 0$ $t = 10, 3$ $α = 10 or 3 β = 3 or 10$ $x^{2} + y^{2} = 10 x^{2} + y^{2} = 3$ $x y = 3 x y = 10$ $(x, y) = (3, 1) & (- 3, - 1) or (1, 3), (- 1, - 3)$
	Answered by York12 last updated on 25/May/23
	$x^2 +y^2 +2xy−2xy=13−xy (x+y)^2 −2xy=13−xy x^4 +3x^2 y^2 +y^4 =((x+y)^2 −2xy_(13−xy) )^2 +x^2 y^2 =109 (13−xy)^2 =109−x^2 y^2 x^2 y^2 −26xy+169=109−x^2 y^2 2x^2 y^2 −26xy+60_(A quadratic equation in (xy)) =0 →(xy−3)(2xy−20)=0 ∴ xy=3 ∨, xy = 10 case (I) for xy = 3 → y = (3/x) substitute (y) in the second equation x^2 +(9/x^2 ) = 10 →x^4 − 10x^2 + 9 = 0 x^2 = 9 → x = +_− ( 3 ) ∨ x^2 =1 → x = +_− (1) then by substituting we get the required values {(3,1) , (−3,−1) ,(1,3) , (−1,−3)}$
	$x^{2} + y^{2} + 2 x y - 2 x y = 13 - x y$ ${(x + y)}^{2} - 2 x y = 13 - x y$ $x^{4} + 3 x^{2} y^{2} + y^{4} = {(\underset{13 - x y}{\underset{⏟}{{(x + y)}^{2} - 2 x y}})}^{2} + x^{2} y^{2} = 109$ ${(13 - x y)}^{2} = 109 - x^{2} y^{2}$ $x^{2} y^{2} - 26 x y + 169 = 109 - x^{2} y^{2}$ $\underset{A q u a d r a t i c e q u a t i o n i n (x y)}{\underset{⏟}{2 x^{2} y^{2} - 26 x y + 60}} = 0 \to (x y - 3) (2 x y - 20) = 0$ $∴ x y = 3 \lor, x y = 10$ $c a s e (I) f o r x y = 3 \to y = \frac{3}{x}$ $s u b s t i t u t e (y) i n t h e s e c o n d e q u a t i o n$ $x^{2} + \frac{9}{x^{2}} = 10 \to x^{4} - 10 x^{2} + 9 = 0$ $x^{2} = 9 \to x = \underline{+} (3) \lor x^{2} = 1 \to x = \underline{+} (1)$ $t h e n b y s u b s t i t u t i n g w e g e t t h e r e q u i r e d v a l u e s$ ${(3, 1), (- 3, - 1), (1, 3), (- 1, - 3)}$
	Answered by witcher3 last updated on 25/May/23
	$2⇒x^4 +y^4 +x^2 y^2 +2x^2 y^2 +2xy(x^2 +y^2 )=169 ⇔2xy(13−xy)=169−109=60 t=xy⇔2t(13−t)−60=0 t^2 −13t+30=(t−10)(t−3) t=3⇒2⇒(x+y)^2 −xy⇒(x+y)^2 =16 x+y=+_− 4 T^2 −4T+3=0⇒(T−3)(T−1)⇒T∈{1,3} (x,y)∈{(1,3);(3,1)} T^2 +4T+3⇒(x,y)∈{(−1,−3),(−3,−1)} x+y=+_− (√(23)) ⇒T^2 −(√(23))T+10=0⇒23−40=−17 (x,y)∈{+_− ((((√(23))+i(√(17)))/2),(((√(23))−i(√(17)))/2))}∈C−R withe if (x,y) Solution (y,x) also$
	$2 \Rightarrow x^{4} + y^{4} + x^{2} y^{2} + {2 x}^{2} y^{2} + 2 xy (x^{2} + y^{2}) = 169$ $\Leftrightarrow 2 xy (13 - xy) = 169 - 109 = 60$ $t = xy \Leftrightarrow 2 t (13 - t) - 60 = 0$ $t^{2} - 13 t + 30 = (t - 10) (t - 3)$ $t = 3 \Rightarrow 2 \Rightarrow {(x + y)}^{2} - xy \Rightarrow {(x + y)}^{2} = 16$ $x + y = \underline{+} 4$ $T^{2} - 4 T + 3 = 0 \Rightarrow (T - 3) (T - 1) \Rightarrow T \in {1, 3}$ $(x, y) \in {(1, 3); (3, 1)}$ $T^{2} + 4 T + 3 \Rightarrow (x, y) \in {(- 1, - 3), (- 3, - 1)}$ $x + y = \underline{+} \sqrt{23}$ $\Rightarrow T^{2} - \sqrt{23} T + 10 = 0 \Rightarrow 23 - 40 = - 17$ $(x, y) \in {\underline{+} (\frac{\sqrt{23} + i \sqrt{17}}{2}, \frac{\sqrt{23} - i \sqrt{17}}{2})} \in C - R$ $withe if (x, y) Solution (y, x) also$
	Commented by senestro last updated on 25/May/23

	$i {d o n}^{'} t u n d e r s t a n d y o u r g e n i u s$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum