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Find all ordered triples of real numbers (x,y,z) that satisfy

 

sqrt(x - y + z) = sqrt(x) - sqrt(y) + sqrt(z)

x + y + z = 8

x - y + z = 4

 Dec 8, 2019
 #1
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From x + y + z = 8 and x - y + z = 4,

\(\color{red}(x+y+z)\color{black}-\color{blue}(x-y+z)\color{black} = \color{red}8\color{black} -\color{blue}4\color{black} = 4\\ 2y = 4\\ y = 2\)
\(x + 2 + z = 8 \\ x + z = 6\)

\(\sqrt{x+z-2}+\sqrt2 = \sqrt x + \sqrt z\\ 2 + \sqrt{2} = \sqrt{x} + \sqrt{z}\\ \text{Let }z = 6-x\\ 2+\sqrt{2} = \sqrt{x} + \sqrt{6 - x}\\ \text{Square both sides,}\\ 6 + 4\sqrt 2 = 6 +2\sqrt{6x-x^2}\\ 2\sqrt 2 = \sqrt{6x - x^2}\\ \text{Square both sides again,}\\ 8 = 6x - x^2\\ x^2 - 6x + 8 = 0\\ (x - 2)(x - 4) = 0\\ x = 2 \text{ or } x = 4\\ \text{When }x = 2,z=4\\ \text{When }x=4,z=2\\ \text{The required ordered triples are }(2,2,4)\text{ and }(4,2,2)\)

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 Dec 8, 2019

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