Let$$y= \frac{x^2 + 34x - 71}{ x^2 + 2x - 7}$$ where $x \in \mathbb R$. Find the range of the given equation.
By cross-multiplying and rearranging the terms, we will get:
$$(1-y)x^2 + (34-2y)x+ (7y-71) = 0 $$
Now, we will put $D\geq 0$ ($x \in \mathbb R$) and then solve the in-equation using wavy curve method to get
$$y \in (-\infty,5] \cup [9,\infty)$$
Now, even though the quadratic equation doesn't remain quadratic for $y=1$ as the coefficient of $x^2$ becomes zero, we get $y=1$ in the range from the same equation.
So, should I make the assumption that $y$ will not be in the range? But we will get $x=2$ when we put $y=1$ in the given rational equation which proves that y=1 is valid.
How should I go about the value $1$ for $y$?
Your analysis is fully correct, what it shows is that for any $y$ there are two solutions for $x$ if the determinant is positive ($y<5$ and $y>9$), no (real) solution for a negative determinant ($5<y<9$), and 1 solution for the boundary cases $y=5$ and $y=9$.
Only at $y=1$ there is something funny going on, because suddenly there is only 1 solution again which is for $x=2$. What you should do here is make a graph of the function $y(x)$ on the interval say $x \in [-10,10]$ and look at the behavior of the curves. Then rotate the graph by 90 degrees and you see how the function $x(y)$ would look like, specially near $y=1$. You will find that there is actually more than one curve, one that behaves nicely near $y=1$ and one that diverges at $y=1$. The latter corresponds with $x = \pm \infty$ and can not be found as a solution to the quadratic equation.
You could also try to determine the function $x(y)$. If you do this correctly, you will find two solutions: $$ x_\pm(y) = \frac{-y+17 \pm 2 \sqrt{2} \sqrt{(y-5)(y-9)}}{y-1} $$ and see that at $y=1$ the solutions are suspicious. If you insert $y=1$ in the solutions you get $x_+(1) = \frac{32}{0}$ which diverges as you would expect. But for the other one you get $x_-(1)=\frac{0}{0}$, which is indeterminate.
If you would do a careful analysis, you would find that for values close to $y=1$, the function does no diverge at all but you get for $y \rightarrow 1$ that $x_-(y) \rightarrow 2$.