# Thread: Factoring to solve an inequality.

1. ## Factoring to solve an inequality.

I know tO factor but I'm not sure here to start
Solve for x:

(16-x^2)/(x^2 +x-6)>or equal to 0

5x^(2)e^x + 2xe^x > 0

Please explain how to solve them, thanks.

2. ## Re: Factoring

Factor the left hand side of each inequality. That's where to start.

3. ## Re: Factoring

Originally Posted by crazycarlos61
I know tO factor but I'm not sure here to start
Solve for x:

(16-x^2)/(x^2 +x-6)>or equal to 0

5x^(2)e^x + 2xe^x > 0

Please explain how to solve them, thanks.
$\displaystyle \displaystyle \frac{16 - x^2}{x^2 + x - 6} \geq 0$

First of all, it should be obvious that $\displaystyle \displaystyle x \neq -3$ and $\displaystyle \displaystyle x \neq 2$. (Why?)

You will need to consider two cases, the first where $\displaystyle \displaystyle x^2 + x - 6 < 0$, and the second where $\displaystyle \displaystyle x^2 + x - 6 > 0$.

Case 1: First note that since the denominator is negative, but the entire fraction is nonnegative, that means the numerator must not be positive. So $\displaystyle \displaystyle 16 - x^2 \leq 0 \implies x^2 - 16 \geq 0$.

\displaystyle \displaystyle \begin{align*}\frac{16 - x^2}{x^2 + x - 6} &\geq 0 \\ 16 - x^2 &\leq 0 \textrm{ (since the denominator is negative, the inequality sign changes)} \\ x^2 - 16 &\geq 0 \\ x^2 &\geq 16 \\ \sqrt{x^2} &\geq \sqrt{16} \\ |x| &\geq 4 \\ x \leq -4 \textrm{ or }x \geq 4 \end{align*}

Now have a try of Case 2.

4. ## Re: Factoring

Ah, thank you to both of you for your help.

But wouldn't case two be incorrect? Seeing as the numerator must be negative to counter the negative denominator to make the result more than zero. If there were a positive numerator wouldn't that mean the equation as a whole I negative an thus less than zero when it's supposed to be greaterthan zero?

5. ## Re: Factoring

Originally Posted by crazycarlos61
I know tO factor but I'm not sure here to start
Solve for x:

(16-x^2)/(x^2 +x-6)>or equal to 0

5x^(2)e^x + 2xe^x > 0

Please explain how to solve them, thanks.
For the second, remember that $\displaystyle \displaystyle e^x > 0$ for all $\displaystyle \displaystyle x$, so

\displaystyle \displaystyle \begin{align*} 5x^2e^x + 2x\,e^x &> 0 \\ e^x\left(5x^2 + 2x\right) &> 0 \\ 5x^2 + 2x &> 0 \\ x^2 + \frac{2}{5}x &> 0 \\ x^2 + \frac{2}{5}x + \left(\frac{1}{5}\right)^2 &> \left(\frac{1}{5}\right)^2 \\ \left(x + \frac{1}{5}\right)^2 &> \frac{1}{25} \\ \left|x + \frac{1}{5}\right| &> \frac{1}{5} \\ x + \frac{1}{5} < -\frac{1}{5} \textrm{ or } x + \frac{1}{5} &> \frac{1}{5} \\ x < -\frac{2}{5} \textrm{ or } x &> 0\end{align*}

6. ## Re: Factoring

You can also make a sign table.

7. ## Re: Factoring

You had me until 2/5x what did you do with that? Also why did you add in (1/5)^2?

8. ## Re: Factoring

Originally Posted by crazycarlos61
You had me until 2/5x what did you do with that? Also why did you add in (1/5)^2?
Completing the Square. If you are dealing with quadratic inequalities then you must have come across Completing the Square to solve quadratic equalities.

9. ## Re: Factoring

Ah okay forgot about that thanks