Wildest Equation

Algebra Level 4

Suppose that $x, y$ and $z$ are real numbers and $y \geq 1$ . Find $\sqrt{x^2 + y^2 +z^2}$ such that

$\large{x+y = 1-2\sqrt{xy-x} - \sqrt{1-2y+y^2 +z^4}}.$

2 5 6 1 3 4

2 solutions

Henry Wan
Oct 22, 2015

$For\quad convenience,\quad let\quad y'=y-1\ge 0,\quad and\quad note\quad that\quad x\ge 0\quad (otherwise,\quad \sqrt { xy-x } is\quad invalid),\\ then\quad the\quad constraint\quad becomes\\ x+y'+2\sqrt { xy' } =-\sqrt { { y' }^{ 2 }+{ z }^{ 4 } } \\ { \left( \sqrt { x } +\sqrt { y' } \right) }^{ 2 }=-\sqrt { { y' }^{ 2 }+{ z }^{ 4 } } \\ Since\quad { \left( \sqrt { x } +\sqrt { y' } \right) }^{ 2 }\ge 0\quad and\quad -\sqrt { { y' }^{ 2 }+{ z }^{ 4 } } \le 0,\quad we\quad have\\ { \left( \sqrt { x } +\sqrt { y' } \right) }^{ 2 }=0\quad and\quad -\sqrt { { y' }^{ 2 }+{ z }^{ 4 } } =0\\ \sqrt { x } +\sqrt { y' } =0\quad and\quad { y' }^{ 2 }+{ z }^{ 4 }\\ x=y'=z=0\\ therefore,\quad x=z=0\quad and\quad y=1\\ The\quad required\quad answer\quad is\quad \boxed { 1 } .$

Bikash Kumar
Oct 15, 2015

x=0 y=1 z=0 I will give solution if anyone need.

Can you show how do you get this value :D

Paul Ryan Longhas - 5 years, 8 months ago

X+2√(x(y-1)) +(y-1) =- √(1-2y+y^2+z^4 ) (√x+√((y-1))) 2+ √(〖(y-1)〗^2+z^4 )=0 For the above statement to be true. Both the terms must be equal to zero. 〖(y-1)〗^2+z^4=0 For the above statement to be true. Both the terms must be equal to zero. (y-1)=0 y=1 z=0 put value of y=1 in √x+√((y-1))) =0 and we will get, x=0 Now we can easily get the value of the sked expression.

Bikash Kumar - 5 years, 8 months ago

Wildest Equation

2 solutions

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