The equation x ^ 2 – 4x + [ x ] + 3 = x [ x ] , where [ x ] denotes...

Given:

${x}^{2} - 4x + \left[x\right] + 3 = x$\left[x\right]

$⇒ {x}^{2} - 4x + 3 = x\left[x\right]$ - \left[x\right]$</p><p>$⇒ \left(x - 1\right)\left(x - 3\right) = \left[x\right]\left(x - 1\right)$</p><p>⇒ x = 1 or$x - 3 = \left[x\right]

If $x - 3 = \left[x\right]$

$⇒ x – \left[x\right] = 3$

$⇒ \left\{x\right\} = 3$

which is not Possible, since \left\{x\right\} \in [ 0, 1 )$.</p><p>Hence,$x = 1$is the only one solution in \left(- \infty , \infty\right)$.