1. ## Solving this inequation

I did have a look at the sticky but I'm still confused.

3x^2 + x - 2 > 0
I got
x > 2/3 and x > -1

For the interval notation I got
(-1, infinity) union (2/3, infinity)
(negative infinity, -1) union (2/3, infinity)

Why is that? :S

2. Originally Posted by Klutz
I did have a look at the sticky but I'm still confused.

3x^2 + x - 2 > 0
I got
x > 2/3 and x > -1

For the interval notation I got
(-1, infinity) union (2/3, infinity)
(negative infinity, -1) union (2/3, infinity)

Why is that? :S
3x^2 + x - 2 = (3x - 2)(x + 1) > 0

the graph of the parabola crosses the x-axis at x = -1 and x = 2/3

now, think about the graph of the parabola (opens upward, correct?) ... where is the graph below the x-axis? where is it above?

3. oh I get it now thanks!

is it required to sketch the graph for this everytime?

4. Originally Posted by Klutz
oh I get it now thanks!

is it required to sketch the graph for this everytime?
not really "required", but it sure helps to visualize the intervals.

5. Hello, Klutz!

You had: . $3x^2 + x - 2 \:>\:0$

You factored and got: . $(x + 1)(3x-2) \:>\:0$

We have the product of two factors and it's suppose to be positive.
. . Under what conditions is this true?

Obviously, if both factors are positive, their product is positive.
. . (You saw this one.) **

Also, if both factors are negative, their product is positive.
. . (You missed this one!)

We have: . $\begin{Bmatrix}x + 1 \:<\:0 & \Rightarrow & x\:<\:-1 \\ 3x-2 \:<\:0 & \Rightarrow & x \:<\:\frac{2}{3} \end{Bmatrix}$

These can be condensed into: . $x \:<\:-1$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

**

You had: . $x \:>\:-1\,\text{ and }\,x \:>\:\frac{2}{3}$

Can you see that this is redundant?

You want numbers greater than $\text{-}1$ and greater than $\frac{2}{3}$

Any number greater than $\frac{2}{3}$ satisfies both inequalities.

They can be condensed: . $x \:>\:\frac{2}{3}$

The answer is: . $\begin{Bmatrix}
x \:<\:\text{-}1 & \text{or} & x \:>\:\frac{2}{3} \\ \\[-2mm]
(\text{-}\infty,\:\text{-}1) & \cup &(\frac{2}{3},\:\infty) \end{Bmatrix}$