Clean the Floor!

Algebra Level 3

⌊ x 3 ⌋ + ⌊ x 2 ⌋ + ⌊ x ⌋ = { x } − 1 \large \lfloor{x^3}\rfloor+\lfloor{x^2}\rfloor+\lfloor{x}\rfloor=\{x\}-1

Solve for the real roots of the equation above.

Notations :


The answer is -1.

A Former Brilliant Member by A Former Brilliant Member

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Firstly, we know that 0 ≤ { x } < 1 0\leq{\{x\}}<1 . The left side of the equation is an integer, implying that the right side is an integer as well, and this further implies that { x } = 0 ⟹ x \{x\}=0\implies{x} is an integer. We may then remove the floor functions to obtain x 3 + x 2 + x + 1 = 0 x^3+x^2+x+1=0 . ( x + 1 ) ( x 2 + 1 ) = 0 (x+1)(x^2+1)=0 ∴ x = − 1 \therefore{x=-1}

3 Helpful 0 Interesting 0 Brilliant 0 Confused

But... what if x not an integer. Will there truly be no other solutions???

Shishir Shahi - 3 years, 11 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...