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anonymous

  • 5 years ago

The value of x? http://img641.imageshack.us/img641/2189/andysnap010.png

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  1. anonymous
    • 5 years ago
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    I answered something before where I explained a solid method for solving this stuff...I'll try and find it. If you learn this method, you won't go wrong.

  2. anonymous
    • 5 years ago
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    http://openstudy.com/groups/mathematics#/updates/4d92db3c0b9d8b0b10a708a9

  3. anonymous
    • 5 years ago
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    Read the part about finding the 'critical points'...near the end.

  4. anonymous
    • 5 years ago
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    The answer's c btw

  5. anonymous
    • 5 years ago
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    oh I'll check it out now

  6. anonymous
    • 5 years ago
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    \[2x^2+x-1=0 \rightarrow x=-1, \frac{1}{2}\]are the points that partition your interval. So you have to take a test point from each of the following intervals \[-\infty < x <-1\]\[-1<x<\frac{1}{2}\]\[\frac{1}{2}<x<\infty\] and test it in the original inequality to see if it's true or false. If it's true, that entire interval has x's that satisfy the inequality. If not, none of the x's in the interval satisfy the inequality. -2, 0 and 1 are good points to use for each interval respectively. Then, for example,\[2(-2)^2+(-2)-1=5>0\]which is true, so the first interval is part of your solution; that is\[\left\{ x:x<-1 \right\}\]

  7. anonymous
    • 5 years ago
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    Oh I'll check it out now

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spraguer (Moderator)
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