I'm working on an exercise from a book in the chapter on quadratic inequalities: "Find the set of possible values of the given function $\frac{x – 2}{(x + 2)(x – 3)}$". The answer in the book is "all values". I don't have much intuition on the calculations so would be grateful if someone could explain what happens at the various stages. Corrections, qualifications, etc. to my statements are very much welcomed!
Here's my workings:
Looking for the range of values of $y$, so $y = \frac{x – 2}{(x + 2)(x – 3)} \Rightarrow y = \frac{x – 2}{x^2 – x – 6}$
$\Rightarrow y(x^2 – x – 6) = x – 2$
$\Rightarrow yx^2 -(y + 1)x – 6y + 2 = 0$
As far as I can understand from my book, now that I've got a quadratic in $x$, I can use the discriminant to determine the $y$ range . With $a = y, b = -(y + 1), c = (-6y + 2)$, the discriminant ($b^2 – 4ac$) is:
$(-y-1)^2 – 4(y)(-6y + 2)$
$= 25y^2 – 6y + 1$
When $25y^2 – 6y + 1 \geqslant 0$, the roots of the quadratic in $x$ are real. At this stage, from what I understand, the values satisfying the inequality $25y^2 – 6y + 1 \geqslant 0$ are the range of $y$ values for all valid, real $x$ values plugged into the function $\frac{x – 2}{(x + 2)(x – 3)}$.
Solving the inequality using the quadratic formula with $a = 25, b = -6, c = 1$:
roots: $\frac{6\pm\sqrt{36 – (4)(25)(1)}}{50}$
$ = \frac{6\pm\sqrt{-64}}{50}$. Because the discriminant here $< 0$ then there are no real roots. I don't get how this is interpreted as "all values" (i.e. the answer in the book).
Best Answer
Since there are no real roots, the graph of $25y^2−6y+1$ cannot cross the horizontal axis and has to be always negative or always positive. Since it is positive for $y=0$, it is always positive.
Since this is the expression for your original equation, this means that you can always find a solution $x$ which gives this value of $y$ (although you should check the trivial fact that this never gives a 0-denominator).