The number of solutions of the equation ${\mathrm{x}}^2+\left[\mathrm{x}\right]-4\mathrm{x}+3=0$ ([.] $\to$denotes the greatest
integer function)

From the given equation, we have  ${\mathrm{x}}^2+(\mathrm{x}-\mathrm{f})-4\mathrm{x}+3=0$ where f= F. P. F. such that $0\le \mathrm{f}<1\Rightarrow \left({\mathrm{x}}^2-3\mathrm{x}+3\right)-\mathrm{f}=0\Rightarrow \mathrm{f}={\mathrm{x}}^2-3\mathrm{x}+3$

But $0\le \mathrm{f}<1$ Therefore  $0\le {\mathrm{x}}^2-3\mathrm{x}+3<1$ Now, solving ${\mathrm{x}}^2-3\mathrm{x}+3=0$ we get  $\mathrm{x}=\frac{3\pm \sqrt{9-12}}{2}=$imaginary $\therefore \mathrm{ }{\mathrm{x}}^2-3\mathrm{x}+3\ge 0$ for all $\mathrm{x}\in \mathrm{R}$ Again from ${\mathrm{x}}^2-3\mathrm{x}+3<1$, we get ${\mathrm{x}}^2-3\mathrm{x}+2<0\Rightarrow (\mathrm{x}-2)(\mathrm{x}-1)<0\Rightarrow \mathrm{ }1<\mathrm{x}<2$Thus the inequality (1) is satisfied if $1<\mathrm{x}<2\Rightarrow \left[\mathrm{x}\right]=1$ 

Putting  $\left[x\right]=1$ in the given equation, we get 

${\mathrm{x}}^2+1-4\mathrm{x}+3=0\Rightarrow {\mathrm{x}}^2-4\mathrm{x}+4=0\Rightarrow (\mathrm{x}-2{)}^2=0\Rightarrow \mathrm{x}=2$But $\mathrm{x}=2$ does not satisfy the inequality   $1<\mathrm{x}<2$

Hence no x is available to satisfy the equation. That is, the given equation has no solution.