Let [x] denote the greatest integer function. What is the number of...

$x^2-4 x+[x]=0$ Given internal, $[0,2]$ Case 1 : Let $0 \leq \mathrm{x} < 1[\mathrm{x}]=0\therefore x^2-4 x+0=0 \Rightarrow x(x-4)=0 \Rightarrow x=0, x=4x=4$ can't be taken in $0 \leq x < 1\therefore \mathrm{x}=0$ Case 2 : Let $1 \leq \mathrm{x} < 2$$\begin{aligned} & {[\mathrm{x}]=1} \\ & \therefore \mathrm{x}^2-4 \mathrm{x}+1=0 \end{aligned}$roots are $\frac{4 \pm \sqrt{16-4}}{2}=\frac{4 \pm \sqrt{12}}{2}=2 \pm \sqrt{3}$ In internal $1 \leq x < 2,2 \pm \sqrt{3}$ are not the roots. Case $3:$ Let $x=2\begin{aligned} & {[\mathrm{x}]=2} \\ & \therefore \mathrm{x}^2-4 \mathrm{x}+2=0 \end{aligned}$$$roots are$\frac{4 \pm \sqrt{16-8}}{2}=\frac{4 \pm \sqrt{8}}{2}=2 \pm \sqrt{2}$Since,$\mathrm{x}=2$, roots can't be$2 \pm \sqrt{2}\therefore$There is only one solution,$\mathrm{x}=0$