If x³ - 4x² + 19 = 6(x - 1) then what is the value of $$\left[ {{x^2} + \frac{1}{{x - 4}}} \right]?$$

Solution(By Examveda Team)

$$\eqalign{ & {x^3} - 4{x^2} + 19 = 6\left( {x - 1} \right)......\left( 1 \right)\left( {{\text{Given}}} \right) \cr & \& \,\left( {{x^2} + \frac{1}{{x - 4}}} \right) = ?\left( {{\text{To find}}} \right) \cr & \to \frac{{{x^2}\left( {x - 4} \right) + 1}}{{x - 4}}......\left( 2 \right) \cr & {\text{from equation}}\left( 1 \right) \cr & {x^3} - 4{x^2} + 19 = 6x - 6 \cr & {x^3} - 4{x^2} + 1 + 18 = 6x - 6 \cr & {x^3} - 4{x^2} + 1 = 6x - 6 - 18 \cr & {x^3} - 4{x^2} + 1 = 6x - 24 \cr & {\text{putting the value of }}\left( {{x^3} - 4{x^2} + 1} \right){\text{in equation }}\left( 2 \right) \cr & {\text{we get,}} = \frac{{6x - 24}}{{x - 4}} = \frac{{6\left( {x - 4} \right)}}{{\left( {x - 4} \right)}} = 6 \cr} $$