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Question Number 129874 by mohammad17 last updated on 20/Jan/21

Commented by mohammad17 last updated on 20/Jan/21

pleas sir help me

$${pleas}\:{sir}\:{help}\:{me} \\ $$

Answered by liberty last updated on 21/Jan/21

((x),(y) ) = (((cos (π/4) −sin (π/4))),((sin (π/4) cos (π/4))) )^(−1) (((x′)),((y′)) ) ((x),(y) ) = (((((√2)/2) −((√2)/2))),((((√2)/2) ((√2)/2))) )^(−1) (((x′)),((y′)) ) ((x),(y) ) = (1/1) ((( ((√2)/2) ((√2)/2))),((−((√2)/2) ((√2)/2))) ) (((x′)),((y′)) ) ((x),(y) ) = ((( ((√2)/2)(x′+y′))),((−((√2)/2)(x′−y′))) ) ⇔(√(((√2)/2)(x+y))) +(√(((√2)/2)(y−x))) = 1 ⇔ (2^(1/4) /2^(1/2) ) ((√(x+y)) +(√(y−x)) )=1 (√(x+y)) + (√(y−x)) = (2)^(1/4)

$$\:\begin{pmatrix}{\mathrm{x}}\\{\mathrm{y}}\end{pmatrix}\:=\:\begin{pmatrix}{\mathrm{cos}\:\frac{\pi}{\mathrm{4}}\:\:\:\:−\mathrm{sin}\:\frac{\pi}{\mathrm{4}}}\\{\mathrm{sin}\:\frac{\pi}{\mathrm{4}}\:\:\:\:\:\:\:\mathrm{cos}\:\frac{\pi}{\mathrm{4}}}\end{pmatrix}^{−\mathrm{1}} \begin{pmatrix}{\mathrm{x}'}\\{\mathrm{y}'}\end{pmatrix} \\ $$$$\begin{pmatrix}{\mathrm{x}}\\{\mathrm{y}}\end{pmatrix}\:=\:\begin{pmatrix}{\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:\:\:\:\:−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}}\\{\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:\:\:\:\:\:\:\:\:\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}}\end{pmatrix}^{−\mathrm{1}} \begin{pmatrix}{\mathrm{x}'}\\{\mathrm{y}'}\end{pmatrix} \\ $$$$\begin{pmatrix}{\mathrm{x}}\\{\mathrm{y}}\end{pmatrix}\:=\:\frac{\mathrm{1}}{\mathrm{1}}\begin{pmatrix}{\:\:\:\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:\:\:\:\:\:\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}}\\{−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:\:\:\:\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}}\end{pmatrix}\:\:\begin{pmatrix}{\mathrm{x}'}\\{\mathrm{y}'}\end{pmatrix} \\ $$$$\begin{pmatrix}{\mathrm{x}}\\{\mathrm{y}}\end{pmatrix}\:=\:\begin{pmatrix}{\:\:\:\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left(\mathrm{x}'+\mathrm{y}'\right)}\\{−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left(\mathrm{x}'−\mathrm{y}'\right)}\end{pmatrix} \\ $$$$\Leftrightarrow\sqrt{\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left(\mathrm{x}+\mathrm{y}\right)}\:+\sqrt{\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left(\mathrm{y}−\mathrm{x}\right)}\:=\:\mathrm{1} \\ $$$$\Leftrightarrow\:\frac{\mathrm{2}^{\mathrm{1}/\mathrm{4}} }{\mathrm{2}^{\mathrm{1}/\mathrm{2}} }\:\left(\sqrt{\mathrm{x}+\mathrm{y}}\:+\sqrt{\mathrm{y}−\mathrm{x}}\:\right)=\mathrm{1} \\ $$$$\:\sqrt{\mathrm{x}+\mathrm{y}}\:+\:\sqrt{\mathrm{y}−\mathrm{x}}\:=\:\sqrt[{\mathrm{4}}]{\mathrm{2}} \\ $$

Commented by MJS_new last updated on 20/Jan/21

error: it must be (√(x+y))+(√(y−x))=(2)^(1/4) (√x)+(√y)=1 ⇔ y=(1−(√x))^2 =1+x−2(√x) defined for y≥0∧x≥0 (√(x+y))+(√(y−x))=(2)^(1/4) ⇔ y=((√2)/4)(2x^2 +1) the endpoint ((0),(1) ) of the first graph becomes (((−((√2)/2))),(((√2)/2)) ) ⇒ defined for x≥−((√2)/2) this is a part of a parabola

$$\mathrm{error}:\:\mathrm{it}\:\mathrm{must}\:\mathrm{be}\:\sqrt{{x}+{y}}+\sqrt{{y}−{x}}=\sqrt[{\mathrm{4}}]{\mathrm{2}} \\ $$$$\sqrt{{x}}+\sqrt{{y}}=\mathrm{1}\:\Leftrightarrow\:{y}=\left(\mathrm{1}−\sqrt{{x}}\right)^{\mathrm{2}} =\mathrm{1}+{x}−\mathrm{2}\sqrt{{x}} \\ $$$$\:\:\:\:\:\mathrm{defined}\:\mathrm{for}\:{y}\geqslant\mathrm{0}\wedge{x}\geqslant\mathrm{0} \\ $$$$\sqrt{{x}+{y}}+\sqrt{{y}−{x}}=\sqrt[{\mathrm{4}}]{\mathrm{2}}\:\Leftrightarrow\:{y}=\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\left(\mathrm{2}{x}^{\mathrm{2}} +\mathrm{1}\right) \\ $$$$\:\:\:\:\:\mathrm{the}\:\mathrm{endpoint}\:\begin{pmatrix}{\mathrm{0}}\\{\mathrm{1}}\end{pmatrix}\:\mathrm{of}\:\mathrm{the}\:\mathrm{first}\:\mathrm{graph} \\ $$$$\:\:\:\:\:\mathrm{becomes}\:\begin{pmatrix}{−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}}\\{\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}}\end{pmatrix}\:\:\Rightarrow\:\mathrm{defined}\:\mathrm{for}\:{x}\geqslant−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$$\mathrm{this}\:\mathrm{is}\:\mathrm{a}\:\mathrm{part}\:\mathrm{of}\:\mathrm{a}\:\mathrm{parabola} \\ $$

Commented by bemath last updated on 21/Jan/21

just typo. (2^(1/4) /2^(1/2) ) = (1/2^(1/4) ) ⇒(√(x+y)) +(√(y−x)) = 2^(1/4)

$$\mathrm{just}\:\mathrm{typo}.\:\frac{\mathrm{2}^{\mathrm{1}/\mathrm{4}} }{\mathrm{2}^{\mathrm{1}/\mathrm{2}} }\:=\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{1}/\mathrm{4}} }\:\Rightarrow\sqrt{\mathrm{x}+\mathrm{y}}\:+\sqrt{\mathrm{y}−\mathrm{x}}\:=\:\mathrm{2}^{\mathrm{1}/\mathrm{4}} \\ $$

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