Sunday, July 31, 2016

Solve for real solution of the system below: [MATH]\left\lfloor{x^3}\right\rfloor+\left\lfloor{x^2}\right\rfloor+\left\lfloor{x}\right\rfloor=\left\{x\right\}-1[/MATH].

Solve for real solution of the system below:

[MATH]\left\lfloor{x^3}\right\rfloor+\left\lfloor{x^2}\right\rfloor+\left\lfloor{x}\right\rfloor=\left\{x\right\}-1[/MATH].

My solution:

First, observe that the LHS of the equality must yield an integer, this tells us the fractional part of $x$ must be a zero, so this turns the whole equation as:

[MATH]x^3+x^2+x=-1\\x^3+x^2+x+1=0\\(x+1)(x^2+1)=0[/MATH]

This implies $x=-1$ is the only real solution to the system.


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