Thread: Inquality with absoloute value in denominator.

1. Inquality with absoloute value in denominator.

6x2+13x-5
------------ < 0
|5x3-10|

I tried to do this with cases, but everything seems wrong, help would be greatly appreciated.

2. Re: Inquality with absoloute value in denominator.

Originally Posted by mikewezyk
6x2+13x-5
------------ < 0
|5x3-10|

I tried to do this with cases, but everything seems wrong, help would be greatly appreciated.
First of all, note that \displaystyle \displaystyle \begin{align*} x \neq \sqrt[3]{2} \end{align*} (why?)

Then, since \displaystyle \displaystyle \begin{align*} \left|5x^3 - 10\right| > 0 \end{align*} for all \displaystyle \displaystyle \begin{align*} x \neq \sqrt[3]{2} \end{align*}, we can multiply both sides of the inequality by it without changing the inequality sign.

\displaystyle \displaystyle \begin{align*} \frac{6x^2 + 13x - 5}{\left|5x^3 - 10\right|} &< 0 \\ 6x^2 + 13x - 5 &< 0 \\ x^2 + \frac{13}{6}x - \frac{5}{6} &< 0 \\ x^2 + \frac{13}{6}x + \left(\frac{13}{12}\right)^2 - \left(\frac{13}{12}\right)^2 - \frac{5}{6} &< 0 \\ \left(x + \frac{13}{12}\right)^2 - \frac{169}{144} - \frac{120}{144} &< 0 \\ \left(x + \frac{13}{12}\right)^2 &< \frac{289}{144} \\ \left| x + \frac{13}{12}\right| &< \frac{17}{12} \\ -\frac{17}{12} < x + \frac{13}{12} &< \frac{17}{12} \\ -\frac{5}{2} < x &< \frac{1}{3} \end{align*}

So the solution to your inequality is \displaystyle \displaystyle \begin{align*} x \in \left( -\frac{5}{2} , \frac{1}{3} \right) \end{align*}.