Solve: 4x^3 +8xy^2 −4x=0 -----(1) 8x^2 y−4y =0 -----(2) simultaneosly i.e find the stationary point Thank you


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Question Number 203525 by Mastermind last updated on 21/Jan/24

$Solve :$ ${4 x}^{3} + {8 xy}^{2} - 4 x = 0 - - - - - (1)$ ${8 x}^{2} y - 4 y = 0 - - - - - (2)$ $simultaneosly i . e find the stationary point$ $Thank you$
	Answered by aleks041103 last updated on 21/Jan/24
	$(1) x=0 or x^2 +2y^2 −1=0 ⇒(1) is { ((x=0)),((x^2 +2y^2 =1)) :} (2) y=0 or x=±1/(√2) ⇒ { ((4x^3 +8xy^2 −4x=0)),((8x^2 −4y=0)) :} ⇔ { ((x=0)),((y=0)) :} or { ((x=0)),((x=±1/(√2))) :} or { ((x^2 +2y^2 =1)),((y=0)) :} or { ((x^2 +2y^2 =1)),((x=±1/(√2))) :} ⇒ { ((4x^3 +8xy^2 −4x=0)),((8x^2 −4y=0)) :}⇔(x,y)=(0,0),(±1,0),(±1/(√2),±1/2)$
	$(1) x = 0 o r x^{2} + 2 y^{2} - 1 = 0$ $\Rightarrow (1) i s {\begin{cases} x = 0 \\ x^{2} + 2 y^{2} = 1 \end{cases}$ $(2) y = 0 o r x = \pm 1 / \sqrt{2}$ $\Rightarrow {\begin{cases} 4 x^{3} + 8 {x y}^{2} - 4 x = 0 \\ 8 x^{2} - 4 y = 0 \end{cases}$ $\Leftrightarrow {\begin{cases} x = 0 \\ y = 0 \end{cases} o r {\begin{cases} x = 0 \\ x = \pm 1 / \sqrt{2} \end{cases} o r {\begin{cases} x^{2} + 2 y^{2} = 1 \\ y = 0 \end{cases} o r {\begin{cases} x^{2} + 2 y^{2} = 1 \\ x = \pm 1 / \sqrt{2} \end{cases}$ $\Rightarrow {\begin{cases} 4 x^{3} + 8 {x y}^{2} - 4 x = 0 \\ 8 x^{2} - 4 y = 0 \end{cases} \Leftrightarrow (x, y) = (0, 0), (\pm 1, 0), (\pm 1 / \sqrt{2}, \pm 1 / 2)$
	Commented by aleks041103 last updated on 21/Jan/24

	Commented by Mastermind last updated on 21/Jan/24

	$Thank you so much man$
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