1. ## inequality

(x-2)*(x+5) < 0

do you just make the < an = and solve there?

I tried expanding doing:

x^2 + 3x - 10

I know (x + 2)(x + 1) = x^2 + 3x + 2 which is similar, so then

(x + 2)(x + 1) - 8 < 0 .. but that doesn't really help me

hmm

2. Originally Posted by flash101
(x-2)*(x+5) < 0

do you just make the < an = and solve there?

I tried expanding doing:

x^2 + 3x - 10

I know (x + 2)(x + 1) = x^2 + 3x + 2 which is similar, so then

(x + 2)(x + 1) - 8 < 0 .. but that doesn't really help me

hmm
you treat the < as an equal sign, then test your results at the end

(x - 2)(x + 5) < 0

=> x - 2 < 0 or x + 5 < 0

=> x < 2 or x < -5

so now we basically think of, or draw a number line, placing the numbers -5 and 2 in there relative positions, and test what regions the original equation holds true for

3. Hello, flash!

$(x-2)(x+5) \:< \:0$
Leave it in factored form . . .

We have: . $(x - 2)(x + 5) \:<\:0$

It says: the product of two numbers is negative.
So one factor must be positive and the other must be negative.
. . There are two possible cases.

[1] . $(x - 2)$ is positive and $(x + 5)$ is negative.

So we have: . $\begin{array}{ccc}x - 2 \:>\:0 & \Rightarrow & x \:> \:2 \\ x+5 \:<\:0 & \Rightarrow & x \:<\:\text{-}5\end{array}$

Hence, $x$ is a number which greater than 2 and less than -5.
. . This is clearly impossible.

[2] . $(x-2)$ is negative and $(x+5)$ is positive.

So we have: . $\begin{array}{ccc}x - 2 \:<\:0 & \Rightarrow & x \:<\:2 \\ x + 5 \:>\:0 & \Rightarrow & x \:>\:\text{-}5\end{array}$

Hence, $x$ is a number which is less than 2 and greater than -5.

This is possible for $x$ between -5 and +2: . $\boxed{-5 \:<\:x\:<\:2}$

4. Flash, remember when you multiply with a negative number, the $>$ will become a $<$ and vice versa.