1. Nonlinear Inequality problem?

Ok what am I missing here?

x2 + 7x + 6 > 0 I have this, to solve I factor,
(x+1)(x+6)>0 then
x+1>0 Solve for zero
-1 -1
(-1,Infinity) Interval Notation.

Here's where im confused....then
x+6>0
-6 -6
x>-6? right? Answer 2. What am I forgetting cause apparently this is wrong and the
(-6,Infinity) is wrong?

How is this true answer for interval notation (-Infinity, -6) U (-1,Infinity)?
If x>-6, All answers should be going right on number line left.

3. Re: Nonlinear Inequality problem?

$x^2 + 7x + 6 = (x + 1)(x+6)$.

This will be larger than 0 if: (x + 1) and (x+6) are both > 0,

or (x+1) and (x+6) are both < 0.

x+1 > 0 when x > -1. x+6 > 0 when x > -6. So as long as x > -1, (x+1)(x+6) > 0.

x + 1 < 0 when x < -1. x + 6 < 0 when x < -6. So (x+1)(x+6) > 0 as long as x < -6.

4. Re: Nonlinear Inequality problem?

Originally Posted by sage1987
Ok what am I missing here?

x2 + 7x + 6 > 0 I have this, to solve I factor,
(x+1)(x+6)>0 then
a simple way ... think about the graph of the resulting parabola and where the graph is above the x-axis.

5. Re: Nonlinear Inequality problem?

Hello, sage1987!

Ok, what am I missing here?

I have this to solve: . $x^2 + 7x + 6 \:>\: 0$

I factor: . $(x+1)(x+6)\:>\:0$

We have: . $(x + 1)(x+ 6) \:>\:0$
. . The product of two numbers is positive.

There are two ways that this can happen.

(1) Both factors are positive: . $\begin{Bmatrix}x+1 \:>\: 0 & \Rightarrow & x \:>\: \text{-}1 \\ x+6 \:>\: 0 & \Rightarrow & x \:>\: \text{-}6 \end{Bmatrix}\;\;\Rightarrow\;\;x \:>\:-1$

(2) Both factors are negative: . $\begin{Bmatrix}x+1\:<\:0 & \Rightarrow & x \:<\:\text{-}1 \\ x+6 \:<\:0 & \Rightarrow & x \:<\:\text{-}6 \end{Bmatrix} \;\;\Rightarrow\;\;x \:<\<-6$

Therefore: . $(\text{-}\infty, \text{-}6) \cup (\text{-}1, \infty)$