Question

Solve for $x:[x{]}^2=-23[x]-132$, where $[.]$ denotes GIF.

• A. $x\in [-12,-11)$
• B. $x\in [-11,-10)$
• C. $x\in [-12,-10)$

Answer 1

The correct option is C $x\in [-12,-10)$

Given: $[x{]}^2=-23[x]-132$
$\Rightarrow [x{]}^2+23[x]+132=0$

Let $\left[x\right]=t$
$\Rightarrow t^2+23t+132=0$
$\Rightarrow t^2+11t+12t+132=0$
$\Rightarrow t(t+11)+12(t+11)=0$
$\Rightarrow (t+11)(t+12)=0$

Either $t=-11$ or $t=-12$

For $t=-11,$
$\left[x\right]=-11$
Apply $\left[x\right]\le x<\left[x\right]+1$
$\therefore -11\le x<-11+1$
$\Rightarrow -11\le x<-10$
$\Rightarrow x\in [-11,-10)$

For $t=-12,$
$\left[x\right]=-12$
Apply $\left[x\right]\le x<\left[x\right]+1$
$\therefore -12\le x<-12+1$
$\Rightarrow -12\le x<-11$
$\Rightarrow x\in [-12,-11)$

Hence, $x\in [-11,-10)\cup [-12,-11)$
$\Rightarrow x\in [-12,-10)$