ok im not sure about this problem

i am supposed to solve the inequality

$\displaystyle (x-6)^{2}(x+8)<0$ the zero's are 6 and -8

so the answer would be $\displaystyle (\infty ,-8)$?

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- Dec 10th 2011, 03:28 PMslapmaxwell1solve the inequality
ok im not sure about this problem

i am supposed to solve the inequality

$\displaystyle (x-6)^{2}(x+8)<0$ the zero's are 6 and -8

so the answer would be $\displaystyle (\infty ,-8)$? - Dec 10th 2011, 03:30 PMskeeterRe: solve the inequality
- Dec 10th 2011, 03:36 PMPlatoRe: solve the inequality
- Dec 10th 2011, 03:38 PMslapmaxwell1Re: solve the inequality
ok thanks....

- Dec 10th 2011, 03:41 PMProve ItRe: solve the inequality
First of all, you should write your solution regions from smallest to largest.

Anyway, to solve $\displaystyle \displaystyle \begin{align*} (x - 6)^2(x + 8) < 0 \end{align*} $ you are correct that you should check the roots of the equation $\displaystyle \displaystyle \begin{align*} (x - 6)^2(x + 8) = 0 \end{align*} $, which are $\displaystyle \displaystyle \begin{align*} -8, 6 \end{align*} $.

The function will only ever stop being negative or positive at the roots. So we need to check each region that are separated by the roots.

When $\displaystyle \displaystyle \begin{align*} x = -9 , (x - 6)^2(x + 8) = -225 < 0 \end{align*} $

So $\displaystyle \displaystyle \begin{align*} x \in (-\infty, -8) \end{align*} $ is acceptable.

When $\displaystyle \displaystyle \begin{align*} x = 0, (x - 6)^2(x + 8) = 288 > 0 \end{align*} $

So $\displaystyle \displaystyle \begin{align*} x \in (-8, 6) \end{align*} $ is not acceptable.

When $\displaystyle \displaystyle \begin{align*} x = 7, (x - 6)^2(x + 8) = 15 > 0 \end{align*} $

So $\displaystyle \displaystyle \begin{align*} x \in (6, \infty) \end{align*} $ is not acceptable.

Therefore, the solution is $\displaystyle \displaystyle \begin{align*} x \in (-\infty, 8) \end{align*} $.