# Math Help - Single Quotient

1. ## Single 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

2. Originally Posted by blueridge
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

$x^2 - 4 = (x + 2)(x - 2)$

So the first problem says:
$\frac{x}{x - 2} + \frac{x}{(x + 2)(x - 2)}$

Can you take it from there?

-Dan

3. ## yes

Yes, but what what must I do to solve question 2 considering that it has a less than symbol?

4. Originally Posted by blueridge
x/(x - 2) + x/(x^2 - 4) < 0
$\frac{x}{x - 2} + \frac{x}{x^2 - 4} < 0$

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

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

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

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:
$( -\infty, -3) \to \frac{x(x + 3)}{(x + 2)(x - 2)} > 0$ (No!)
$( -3, -2) \to \frac{x(x + 3)}{(x + 2)(x - 2)} < 0$ (Check!)
$( -2, 0) \to \frac{x(x + 3)}{(x + 2)(x - 2)} > 0$ (No!)
$( 0, 2) \to \frac{x(x + 3)}{(x + 2)(x - 2)} < 0$ (Check!)
$( 2, \infty ) \to \frac{x(x + 3)}{(x + 2)(x - 2)} > 0$ (No!)

So our solution set is $(-3, -2) \cup (0, 2)$.

You can verify this on the graph attached below.

-Dan

5. ## Tell 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?

6. Originally Posted by blueridge
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?

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

7. ## ok

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?

8. Originally Posted by blueridge
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?
yes. same approach, just select the appropriate intervals