Question

The solution set of the equation[n²] +[n + 1] -3 = 0 (where fl denotes greatest integer function) i​

Answer 1

$\rm\Huge\underline{\text{Main idea}}$

For $\rm n$ in $\rm k\leq n<k+1$ for an integer $\rm k$, $\rm[n]=k$.

Or,

$\rm[n]=k\iff k\leq n<k+1$

$\rm\Huge\underline{\text{Solution}}$

$\rm\large\underline{\text{Solution A: Floor function}}$

Given equation is,

$\rm[n]^2+[n+1]-3=0$

For $\rm n$ in $\rm k\leq n<k+1$, $\rm k$ is an integer,

$\rm k^2+(k+1)-3=0$

$\rm k^2+k-2=0$

$\rm(k-1)(k+2)=0$

$\rm k=1\textrm{ or }k=-2$

Since $\rm k\leq n<k+1$,

$\rm-2\leq n<-1\textrm{ or }1\leq n<2$

Hence the solution set is $\red{\underline{\rm-2\leq n<-1\textrm{ or }1\leq n<2}\tiny\text{//}}$.

$\rm\large\underline{\text{Solution B: Substitution}}$

Here is the substitution method. The main idea is the same.

Let $\rm k=[n]$. Then,

$\rm k^2+(k+1)-3=0$

$\rm k^2+k-2=0$

$\rm (k-1)(k+2)=0$

$\rm k=1\textrm{ or }k=-2$

Since $\rm k=[n]$,

$\rm[n]=1\textrm{ or }[n]=-2$

$\rm-2\leq n<-1\textrm{ or }1\leq n<2$

Hence the solution set is $\red{\underline{\rm-2\leq n<-1\textrm{ or }1\leq n<2}\tiny\text{//}}$.

$\rm\Huge\underline{\text{Learn more}}$

$\rm[x+\alpha]=[x]+\alpha$

$\alpha$ is an integer.

$\rm[n]=k\iff k\leq n<k+1$

The symbol $\iff$ stands for if and only if, which means the conditions are equivalent.