1. ## Inequality Help

4x^2 - x^3 - 4x
------------------- is > or = to 0
4x^2 - 1

i factored it to

x(x-2)(x-2)
------------- is > or = to 0
(2x-1)(2x+1)

i think

Solving Inequalities: An Overview

this is how to do it, but i can't figure out everything they're doing or how to apply it

thx

2. Check your factorising, there's a -x^3 term on top but after the factorisation it's +ve.

3. Originally Posted by guitarplaya08
4x^2 - x^3 - 4x
------------------- is > or = to 0
4x^2 - 1

i factored it to

x(x-2)(x-2)
------------- is > or = to 0
(2x-1)(2x+1)

thx

$=\frac{-x(x-2)(x-2)}{(2x-1)(2x+1)}\geq0$

$-x(x-2)(x-2)\geq0,\;x\neq\frac{1}{2},\frac{-1}{2}$

$x(x-2)(x-2)\leq0,\;x\neq\frac{1}{2},\frac{-1}{2}$

$x(x-2)^2\leq0,\;x\neq\frac{1}{2},\frac{-1}{2}$

Since $x(x-2)^2\geq0\; always$

So, $x\leq0,\;x\neq\frac{1}{2},\frac{-1}{2}$

4. Originally Posted by Shyam

Since $x(x-2)^2\geq0\; always$
That has to be wrong, check it why.

-----

Here's what I'd do:

\begin{aligned}
\frac{4x^{2}-x^{3}-4x}{4x^{2}-1}&\ge& 0 \\
\frac{x^{3}-4x^{2}+4x}{4x^{2}-1}&\le& 0 \\
\frac{x(x-2)^{2}}{(2x+1)(2x-1)}&\le& 0.
\end{aligned}

Since $(x-2)^2\ge0,$ we're gonna have to worry about the another factors, so, make the following table:

$\begin{array}{*{20}c}{}&\vline& {\left({ - \infty , - \dfrac{1}
{2}} \right)}&\vline&{\left( { - \dfrac{1}
{2},0} \right)}&\vline&{\left( {0,\dfrac{1}
{2}} \right)}&\vline &{\left( {\dfrac{1}
{2},\infty }\right)} \\\hline x &\vline&-&\vline &-&\vline &+&\vline&+\\
\hline{2x + 1}&\vline &-&\vline &+&\vline &+&\vline &+\\
\hline{2x - 1}&\vline &-&\vline &-&\vline &-&\vline &+\\
\hline{}&\vline &-&\vline &+&\vline &-&\vline&+\\\end{array}$

Now, look at where the original expression is negative, hence, the solution set is $S=\left] -\infty ,-\frac{1}{2} \right[\cup \left[ 0,\frac{1}{2} \right[\cup \{2\}.$