(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
Hello, flash!
Leave it in factored form . . .$\displaystyle (x-2)(x+5) \:< \:0$
We have: .$\displaystyle (x - 2)(x + 5) \:<\:0$
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] .$\displaystyle (x - 2)$ is positive and $\displaystyle (x + 5)$ is negative.
So we have: .$\displaystyle \begin{array}{ccc}x - 2 \:>\:0 & \Rightarrow & x \:> \:2 \\ x+5 \:<\:0 & \Rightarrow & x \:<\:\text{-}5\end{array}$
Hence, $\displaystyle x$ is a number which greater than 2 and less than -5.
. . This is clearly impossible.
[2] .$\displaystyle (x-2)$ is negative and $\displaystyle (x+5)$ is positive.
So we have: .$\displaystyle \begin{array}{ccc}x - 2 \:<\:0 & \Rightarrow & x \:<\:2 \\ x + 5 \:>\:0 & \Rightarrow & x \:>\:\text{-}5\end{array}$
Hence, $\displaystyle x$ is a number which is less than 2 and greater than -5.
This is possible for $\displaystyle x$ between -5 and +2: .$\displaystyle \boxed{-5 \:<\:x\:<\:2}$