x-2-x-x-x-0-x-

Question Number 7000 by FilupSmith last updated on 05/Aug/16

x^{2} = x + \sqrt{x}

x = 0 \lor x = ? ?

Commented by Yozzii last updated on 05/Aug/16

L e t u = \sqrt{x} \Rightarrow u^{2} = x \Rightarrow u^{4} = x^{2}

u^{4} = u^{2} + u

u^{4} - u^{2} - u = 0

u (u^{3} - u - 1) = 0

\Rightarrow u = 0 o r u^{3} - u - 1 = 0

u (u^{2} - 1) = 1

u (u - 1) (u + 1) = 1 > 0

\Rightarrow u > 1

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