Menu Close

Solve-for-real-numbers-x-32-x-16-y-2-2-2-x-12-y-

Tinku Tara 0 Comments
Pin




Question Number 158275 by HongKing last updated on 01/Nov/21
Solve for real numbers: x^(32) + x^(16) + y^2 = 2 (√2) x^(12) y
$$\mathrm{Solve}\:\mathrm{for}\:\mathrm{real}\:\mathrm{numbers}: \\ $$$$\mathrm{x}^{\mathrm{32}} \:+\:\mathrm{x}^{\mathrm{16}} \:+\:\mathrm{y}^{\mathrm{2}} \:=\:\mathrm{2}\:\sqrt{\mathrm{2}}\:\mathrm{x}^{\mathrm{12}} \:\mathrm{y} \\ $$$$ \\ $$
Answered by mindispower last updated on 01/Nov/21
x^(32) +x^(16) ≥2x^(24) equalite x^(16) =x^8 ⇒x∈{0,1,−1 2x^(24) +y^2 ≥2(√2)x^(12) y ⇔((√2)x^(12) −y)^2 ≥0 y=(√2)x^(12) ∈{0,(√2)} (x,y)={(0,0);(0,(√2));(1,0);(1,(√2));(−1,0);(−1,(√2))}
$${x}^{\mathrm{32}} +{x}^{\mathrm{16}} \geqslant\mathrm{2}{x}^{\mathrm{24}} \:{equalite}\:{x}^{\mathrm{16}} ={x}^{\mathrm{8}} \Rightarrow{x}\in\left\{\mathrm{0},\mathrm{1},−\mathrm{1}\right. \\ $$$$\mathrm{2}{x}^{\mathrm{24}} +{y}^{\mathrm{2}} \geqslant\mathrm{2}\sqrt{\mathrm{2}}{x}^{\mathrm{12}} {y} \\ $$$$\Leftrightarrow\left(\sqrt{\mathrm{2}}{x}^{\mathrm{12}} −{y}\right)^{\mathrm{2}} \geqslant\mathrm{0} \\ $$$${y}=\sqrt{\mathrm{2}}{x}^{\mathrm{12}} \in\left\{\mathrm{0},\sqrt{\mathrm{2}}\right\} \\ $$$$\left({x},{y}\right)=\left\{\left(\mathrm{0},\mathrm{0}\right);\left(\mathrm{0},\sqrt{\mathrm{2}}\right);\left(\mathrm{1},\mathrm{0}\right);\left(\mathrm{1},\sqrt{\mathrm{2}}\right);\left(−\mathrm{1},\mathrm{0}\right);\left(−\mathrm{1},\sqrt{\mathrm{2}}\right)\right\} \\ $$$$ \\ $$
Commented by HongKing last updated on 02/Nov/21
cool my dear Ser, thank you
$$\mathrm{cool}\:\mathrm{my}\:\mathrm{dear}\:\mathrm{Ser},\:\mathrm{thank}\:\mathrm{you} \\ $$
Answered by MJS_new last updated on 02/Nov/21
we can easily solve for y: y=x^8 ((√2)x^4 +∣x^8 −1∣i) ⇒ to stay in R we need x^8 =0 ∨ x^8 −1 =0 ⇒ ⇒ x=0 ∨ x=±1 ⇒ x=0∧y=0 ∨ x=±1∧y=(√2)
$$\mathrm{we}\:\mathrm{can}\:\mathrm{easily}\:\mathrm{solve}\:\mathrm{for}\:{y}: \\ $$$${y}={x}^{\mathrm{8}} \left(\sqrt{\mathrm{2}}{x}^{\mathrm{4}} +\mid{x}^{\mathrm{8}} −\mathrm{1}\mid\mathrm{i}\right) \\ $$$$\Rightarrow\:\mathrm{to}\:\mathrm{stay}\:\mathrm{in}\:\mathbb{R}\:\mathrm{we}\:\mathrm{need}\:{x}^{\mathrm{8}} =\mathrm{0}\:\vee\:{x}^{\mathrm{8}} −\mathrm{1}\:=\mathrm{0}\:\Rightarrow \\ $$$$\Rightarrow\:{x}=\mathrm{0}\:\vee\:{x}=\pm\mathrm{1} \\ $$$$\Rightarrow \\ $$$${x}=\mathrm{0}\wedge{y}=\mathrm{0}\:\vee\:{x}=\pm\mathrm{1}\wedge{y}=\sqrt{\mathrm{2}} \\ $$
Commented by HongKing last updated on 02/Nov/21
cool my dear Sir, thank you so much
$$\mathrm{cool}\:\mathrm{my}\:\mathrm{dear}\:\boldsymbol{\mathrm{S}}\mathrm{ir},\:\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *