Solve : x² - 6x +[x] +7 =0 where [ ] represents greatest integer function

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Please find below the solution to the asked query​

$$\begin{gathered} \mathrm{Given} \mathrm{equation} \mathrm{is} \\ {x}^2-6x+\left[x\right]+7=0 \end{gathered}$$
$\Rightarrow$$x^2-5x-x+\left[x\right]+7=0$
$\Rightarrow$$x^2-5x+7=x-\left[x\right]$
$$\begin{gathered} \mathrm{Now} \\ x-\left[x\right]=\left\{x\right\}; \left\{\mathrm{where} \left\{x\right\} \mathrm{denotes} \mathrm{denotes} \mathrm{fractional} \mathrm{part} \mathrm{of} \mathrm{x} \mathrm{and} \mathrm{range} \mathrm{of} \left\{x\right\} \mathrm{is} [0, 1) \right\} \end{gathered}$$

$\Rightarrow x^2-5x+7=\left\{x\right\}$

$$\begin{gathered} \mathrm{The} \mathrm{above} \mathrm{equation} \mathrm{can} \mathrm{easily} \mathrm{be} \mathrm{solved} \mathrm{graphically}. \mathrm{Let} \\ g\left(x\right)=\left\{x\right\} \\ \mathrm{and} \end{gathered}$$ 
$f\left(x\right)=x^2-5x+7 \mathrm{which} \mathrm{is} \mathrm{an} \mathrm{equation} \mathrm{of} \mathrm{vertical} \mathrm{parabola}$

$\mathrm{So} \mathrm{we} \mathrm{have} \mathrm{to} \mathrm{solve} f\left(x\right)=g\left(x\right)$
 $$\begin{gathered} \mathrm{Consider} a{x}^2+bx+c=0 \\ \mathrm{Vertex} \mathrm{of} \mathrm{above} \mathrm{equation} \mathrm{is} \left\{-b/2a, -\left(b^2-4ac\right)/4a\right\} \end{gathered}$$

$$\begin{gathered} \Rightarrow \mathrm{Vertex} \mathrm{of} f\left(x\right)= \left\{-\left(-5\right)/2, -\left(25-4\times 7\times 1\right)/4\right\}= \left(2.5, 0.75\right) \\ f\left(2\right)=2\times 2-5\times 2+7= 1 \mathrm{and} f\left(3\right)=3\times 3-5\times 3+7= 1 \\ \mathrm{Let} \mathrm{us} \mathrm{now} \mathrm{draw} \mathrm{the} \mathrm{graph} \mathrm{of} f\left(x\right) \mathrm{and} g\left(x\right) \end{gathered}$$



$\mathrm{From} \mathrm{above} \mathrm{figure}\left(1\right) \mathrm{it} \mathrm{is} \mathrm{clear} \mathrm{that} f\left(x\right) \mathrm{and} g\left(x\right) \mathrm{do} \mathrm{not} \mathrm{intersect} \mathrm{each} \mathrm{other} \mathrm{which} \mathrm{means} f\left(x\right) \mathrm{is} \mathrm{never} \mathrm{equal} \mathrm{to} g\left(x\right) ; \left\{\mathrm{Note}: \left\{x\right\}=0 \mathrm{when} \mathrm{x} \mathrm{is} \mathrm{an} \mathrm{integer} .\mathrm{Hence} g\left(2\right)=g\left(3\right)=0 \mathrm{whereas} f\left(2\right)=f\left(3\right)=1\right\}$

$\Rightarrow$$x^2-5x+7=\left\{x\right\} \mathrm{will} \mathrm{hold} \mathrm{for} \mathrm{no} \mathrm{value} \mathrm{of} \mathrm{x}$
 $\mathrm{Therefore} \mathrm{the} \mathrm{given} \mathrm{equation} x^2-6x+\left[x\right]+7=0 \mathrm{has} \mathrm{no} \mathrm{solution}.$

$\Rightarrow x\in \varnothing$

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