Quadratic Discriminant

Consider $\frac { { x }^{ 2 }+2x-11 }{ x-3 } =y$ Then we have, $\quad { x }^{ 2 }+2x-11=xy-3y\\ \Rightarrow { x }^{ 2 }+(2-y)x-11+3y=0..........(i)$ Now, since we have to find the range of $y$ where $x$ is real, we consider the discriminant ${{ (2-y) }^{ 2 }-4.1.(3y-11)}$ of equation (i) and observe that it should be greater than or equal to zero since equation (i) has to give real solutions for $x$ . Thus ${ (2-y) }^{ 2 }-4.1.(3y-11)\ge 0\\ \Rightarrow { y }^{ 2 }-16y+48\ge 0\\ \Rightarrow (y-4)(y-12)\ge 0\\ \Rightarrow y\le 4,\quad or,\quad y\ge 12\\ \Rightarrow y\notin (4,12)$ Hence the result.

We first rewrite the numerator as $\LARGE\frac{x^{2} + 2( x -3) -5}{x-3}$ = $\LARGE 2$ + $\LARGE \frac{x^{2} -5}{x-3}$

Let $\LARGE \frac{x^{2} -5}{x-3}$ = $\LARGE y$

Now proceed along the lines of what Kuldeep has done. Just remember that 2 must be added to lower and upper limits obtained from this expression.