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Math Help - Inequality Help

  1. #1
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    Inequality Help

    Im stuck on this.



    I factorised it to (x-3)(x-2)

    so there for

    x = 3
    x = 2

    Im not sure how to find the values for which it is smaller/equal to 0.

    Any suggestions?
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  2. #2
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    Now you know where it crosses the x-axis, if you draw a quick sketch of the graph, you'll be able to see easily where the values you want lie.
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  3. #3
    o_O
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    Pick numbers in the intervals (-\infty, 2), (2,3), (3, \infty) and test them to see if any of them satisfies your inequality. The one that does is representative of all the numbers on that interval.
    Code:
     
     
    <---------|---------|--------->
              2         3
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  4. #4
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    Quote Originally Posted by housefire View Post
    Now you know where it crosses the x-axis, if you draw a quick sketch of the graph, you'll be able to see easily where the values you want lie.
    Yeh thanks that works. I got it as between 2 and 3, however that does depend on whether the quadratic has its turning point at the top or the bottom. Is there any way to tell from the equation?
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  5. #5
    Moo
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    Hello,

    If you have two numbers a and b, the product ab is negative if and only if a and b are of different signs.

    This means that if a<0, b>0 and if a>0, b<0
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  6. #6
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    \begin{aligned}<br />
   x^{2}-5x+6&\le 0 \\ <br />
  4x^{2}-20x+24&\le 0 \\ <br />
  (2x-5)^{2}&\le 1 \\ <br />
  \left| 2x-5 \right|&\le 1,<br />
\end{aligned}

    hence -1\le 2x-5\le 1\implies 4\le 2x\le 6\,\therefore \,2\le x\le 3.
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