{ ((x^2 +y(√(xy)) = 72)),((y^2 +x(√(xy)) = 36)) :}


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 154099 by EDWIN88 last updated on 14/Sep/21
${ ((x^2 +y(√(xy)) = 72)),((y^2 +x(√(xy)) = 36)) :}$
${\begin{cases} x^{2} + y \sqrt{x y} = 72 \\ y^{2} + x \sqrt{x y} = 36 \end{cases}$
	Answered by EDWIN88 last updated on 14/Sep/21
	${ ((x^2 +y(√(xy)) =72)),((2y^2 +2x(√(xy)) =72)) :} ⇒x^2 −2y^2 +(√(xy))(y−2x)=0 ⇒(√(xy))(y−2x)=2y^2 −x^2 ⇒xy(y^2 −4xy+4x^2 )=4y^4 −4x^2 y^2 +x^4 ⇒xy^3 −4x^2 y^2 +4x^3 y=4y^4 −4x^2 y^2 +x^4 ⇒xy^3 +4x^3 y−4y^4 −x^4 =0 ⇒x(y^3 −x^3 )+4y(x^3 −y^3 )=0 ⇒x(y^3 −x^3 )−4y(y^3 −x^3 )=0 ⇒(y^3 −x^3 )(x−4y)=0 case(1)x=4y { ((16y^2 +y∣2y∣=72)),((y^2 +4y∣2y∣=36)) :} it defined for y≥0 ⇒ { ((18y^2 =72)),((9y^2 =36)) :}⇒y=2 ∧x=8 case(2)y=x { ((x^2 +x∣x∣=72)),((x^2 +x∣x∣=36)) :} inconsisten has no solution for x=y ∴ solution is (x,y)=(8,2)$
	${\begin{cases} x^{2} + y \sqrt{x y} = 72 \\ 2 y^{2} + 2 x \sqrt{x y} = 72 \end{cases}$ $\Rightarrow x^{2} - 2 y^{2} + \sqrt{x y} (y - 2 x) = 0$ $\Rightarrow \sqrt{x y} (y - 2 x) = 2 y^{2} - x^{2}$ $\Rightarrow x y (y^{2} - 4 x y + 4 x^{2}) = 4 y^{4} - 4 x^{2} y^{2} + x^{4}$ $\Rightarrow {x y}^{3} - 4 x^{2} y^{2} + 4 x^{3} y = 4 y^{4} - 4 x^{2} y^{2} + x^{4}$ $\Rightarrow {x y}^{3} + 4 x^{3} y - 4 y^{4} - x^{4} = 0$ $\Rightarrow x (y^{3} - x^{3}) + 4 y (x^{3} - y^{3}) = 0$ $\Rightarrow x (y^{3} - x^{3}) - 4 y (y^{3} - x^{3}) = 0$ $\Rightarrow (y^{3} - x^{3}) (x - 4 y) = 0$ $c a s e (1) x = 4 y$ ${\begin{cases} 16 y^{2} + y ∣ 2 y ∣= 72 \\ y^{2} + 4 y ∣ 2 y ∣= 36 \end{cases}$ $i t d e f i n e d f o r y ⩾ 0$ $\Rightarrow {\begin{cases} 18 y^{2} = 72 \\ 9 y^{2} = 36 \end{cases} \Rightarrow y = 2 \land x = 8$ $c a s e (2) y = x$ ${\begin{cases} x^{2} + x ∣ x ∣= 72 \\ x^{2} + x ∣ x ∣= 36 \end{cases} i n c o n s i s t e n$ $h a s n o s o l u t i o n f o r x = y$ $∴ s o l u t i o n i s (x, y) = (8, 2)$
	Commented by Tawa11 last updated on 15/Sep/21

	$nice$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum