Jul 4, 2012 Thread starter #1 J jacks Well-known member Apr 5, 2012 226 Solve the equation $\mid x \mid+\sqrt{\lfloor x \rfloor+\sqrt{1+\{x\}}} = 1$ where $\lfloor x \rfloor = $ floor function

Solve the equation $\mid x \mid+\sqrt{\lfloor x \rfloor+\sqrt{1+\{x\}}} = 1$ where $\lfloor x \rfloor = $ floor function

Jul 4, 2012 #2 E Evgeny.Makarov Well-known member MHB Math Scholar Jan 30, 2012 2,492 Obviously, |x| <= 1, and x = 0 is one solution. Further, {x} >= 0 for all x, so x cannot be positive. If -1 <= x < 0, then |x| = -x, floor(x) = -1 and {x} = x + 1. This gives an equation that has two solutions: one simple and one complicated.

Obviously, |x| <= 1, and x = 0 is one solution. Further, {x} >= 0 for all x, so x cannot be positive. If -1 <= x < 0, then |x| = -x, floor(x) = -1 and {x} = x + 1. This gives an equation that has two solutions: one simple and one complicated.