Math, asked by lambasheetal, 5 months ago

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

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Answered by TakenName
8

\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.

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