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Math Help - Inequality Help

  1. #1
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    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
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  2. #2
    Super Member Matt Westwood's Avatar
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    Check your factorising, there's a -x^3 term on top but after the factorisation it's +ve.
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  3. #3
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    Quote Originally Posted by guitarplaya08 View Post
    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}
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  4. #4
    Math Engineering Student
    Krizalid's Avatar
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    Quote Originally Posted by Shyam View Post

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

    -----

    Here's what I'd do:

    \begin{aligned}<br />
   \frac{4x^{2}-x^{3}-4x}{4x^{2}-1}&\ge& 0 \\ <br />
  \frac{x^{3}-4x^{2}+4x}{4x^{2}-1}&\le& 0 \\ <br />
  \frac{x(x-2)^{2}}{(2x+1)(2x-1)}&\le& 0.<br />
\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}<br />
{2}} \right)}&\vline&{\left( { - \dfrac{1}<br />
{2},0} \right)}&\vline&{\left( {0,\dfrac{1}<br />
{2}} \right)}&\vline &{\left( {\dfrac{1}<br />
{2},\infty }\right)}  \\\hline x &\vline&-&\vline &-&\vline &+&\vline&+\\<br />
\hline{2x + 1}&\vline &-&\vline &+&\vline &+&\vline &+\\<br />
\hline{2x - 1}&\vline &-&\vline &-&\vline &-&\vline &+\\<br />
\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\}.
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