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

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- Jun 10th 2007, 04:10 AMblueridgeSingle Quotient
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 - Jun 10th 2007, 04:53 AMtopsquark
- Jun 10th 2007, 05:45 AMblueridgeyes
Yes, but what what must I do to solve question 2 considering that it has a less than symbol?

- Jun 10th 2007, 08:06 AMtopsquark
$\displaystyle \frac{x}{x - 2} + \frac{x}{x^2 - 4} < 0$

$\displaystyle \frac{x(x + 2) + x}{(x + 2)(x - 2)} < 0$

$\displaystyle \frac{x^2 + 3x}{(x + 2)(x - 2)} < 0$

$\displaystyle \frac{x(x + 3)}{(x + 2)(x - 2)} < 0$

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 - Jun 10th 2007, 11:13 AMblueridgeTell me...
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? - Jun 10th 2007, 11:17 AMtopsquark
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 - Jun 10th 2007, 12:28 PMblueridgeok
So, basically I approach the same question in terms of other inequality symbols in like manner, right?

The main idea is to select points from various intervals, right? - Jun 10th 2007, 12:31 PMJhevon