I don't really understand how to solve non-linear inequalities after the factoring step? I don't understand how to use sign charts!

For example,

(x-5)(x+4)>=0

or

x^2>=9

???

Please and thank you!

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- Sep 22nd 2012, 03:56 PMsamiileigh94Non-Linear Inequalities, Sign Charts?
I don't really understand how to solve non-linear inequalities after the factoring step? I don't understand how to use sign charts!

For example,

(x-5)(x+4)>=0

or

x^2>=9

???

Please and thank you! - Sep 22nd 2012, 05:41 PMProve ItRe: Non-Linear Inequalities, Sign Charts?
Well, if you think about where the function actually equals 0, it has to be at the x intercepts. Surely you can figure those out.

Then, we make note of the fact that a function can only ever change from positive to negative, or vice versa, AT the x-intercepts.

Therefore, the x intercepts break up the domain into a number of regions where the function could be positive or could be negative.

Test each of those regions by substituting a point from each of those regions into the function. See if you get positive or negative.

The solution to each inequality should then easily follow. - Sep 22nd 2012, 05:48 PMsamiileigh94Re: Non-Linear Inequalities, Sign Charts?
Well I understand the part you've explained, so let me be clearer. I don't understand how the positive and negative changes/leads to the solution. Like, what does it lend to the solution if its positive or negative? If that isn't clear enough, I'd appreciate explanation with an example.

- Sep 22nd 2012, 05:53 PMProve ItRe: Non-Linear Inequalities, Sign Charts?
In the first example you are asked to solve $\displaystyle \displaystyle \begin{align*} (x - 5)( x + 4) \geq 0 \end{align*}$. So you are asked to find all the points where this function is either 0 or positive. Finding where the function is 0 should be easy. Finding where the function is positive is done by testing points in each region determined by the intercepts (where the function is 0).

- Sep 22nd 2012, 06:01 PMsamiileigh94Re: Non-Linear Inequalities, Sign Charts?
So the solution would be x>=-4, x>=5?

- Sep 22nd 2012, 06:06 PMProve ItRe: Non-Linear Inequalities, Sign Charts?
No. Like I said, test points in each region. The first region is $\displaystyle \displaystyle \begin{align*} (-\infty, -4) \end{align*}$ so you could test say $\displaystyle \displaystyle \begin{align*} x = -5 \end{align*}$, the second region is $\displaystyle \displaystyle \begin{align*} (-4, 5) \end{align*}$ so you could test say $\displaystyle \displaystyle \begin{align*} x = 0 \end{align*}$, and the final region is $\displaystyle \displaystyle \begin{align*} (5, \infty) \end{align*}$ so you could test say $\displaystyle \displaystyle \begin{align*} x = 6 \end{align*}$.

- Sep 22nd 2012, 06:18 PMsamiileigh94Re: Non-Linear Inequalities, Sign Charts?
Okay, I get that, but the whole time I've been saying I don't understand how to interpret the results!

I'm attaching a copy of the sign chart for that example. Please help me understand it. - Sep 22nd 2012, 06:23 PMProve ItRe: Non-Linear Inequalities, Sign Charts?
OK, you have just shown that when $\displaystyle \displaystyle \begin{align*} x = -5, (x - 5)(x + 4) > 0 \end{align*}$. That means that ALL POINTS in the region $\displaystyle \displaystyle \begin{align*} (-\infty, -4) \end{align*}$ will be positive. You have that when $\displaystyle \displaystyle \begin{align*} x = 0, (x - 5)(x + 4) < 0 \end{align*}$, so ALL POINTS in the region $\displaystyle \displaystyle \begin{align*} (-4, 5) \end{align*}$ will be negative. Finally you have that when $\displaystyle \displaystyle \begin{align*} x = 6, (x - 5)(x + 4) > 0 \end{align*}$, so ALL POINTS in the region $\displaystyle \displaystyle \begin{align*} (5, \infty) \end{align*}$ will be positive.

Your questions asks "at what points is $\displaystyle \displaystyle \begin{align*} (x - 5)(x + 4) \geq 0 \end{align*}$?"

You know where it equals 0, you know where it is positive. Put those answers together.