Write each expression as a single quotient.
What does the author mean by single quotient?
(1) x/(x - 2) + x/(x^2 - 4)
(2) x/(x - 2) + x/(x^2 - 4) < 0
$\displaystyle \frac{x}{x - 2} + \frac{x}{x^2 - 4} < 0$
$\displaystyle \frac{x(x + 2) + x}{(x + 2)(x - 2)} < 0$
$\displaystyle \frac{x^2 + 3x}{(x + 2)(x - 2)} < 0$
$\displaystyle \frac{x(x + 3)}{(x + 2)(x - 2)} < 0$
Critical numbers are numbers where either the numerator or denominator is 0. In this case the critical numbers are -3, -2, 0, and 2.
So break up the real number line into 5 intervals and test the inequality on each interval:
$\displaystyle ( -\infty, -3) \to \frac{x(x + 3)}{(x + 2)(x - 2)} > 0$ (No!)
$\displaystyle ( -3, -2) \to \frac{x(x + 3)}{(x + 2)(x - 2)} < 0$ (Check!)
$\displaystyle ( -2, 0) \to \frac{x(x + 3)}{(x + 2)(x - 2)} > 0$ (No!)
$\displaystyle ( 0, 2) \to \frac{x(x + 3)}{(x + 2)(x - 2)} < 0$ (Check!)
$\displaystyle ( 2, \infty ) \to \frac{x(x + 3)}{(x + 2)(x - 2)} > 0$ (No!)
So our solution set is $\displaystyle (-3, -2) \cup (0, 2)$.
You can verify this on the graph attached below.
-Dan
Is x/(x - 2) + x/(x^2 - 4) < 0 called a quadratic inequality?
What about if the symbol is the greater than (>) applied to the same question?
What about if the symbol is "greater than or equal to" or if the symbol is "less than or equal to"?
What would I do in those particular cases?
I don't know about the name, but quadratic inequality sounds good.
As for what you do in the other cases, consider that there is an automatic restriction that the denominator not be zero, but depending on the type of inequality critical points that make the numerator zero may be allowed.
-Dan