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Math Help - Help with solving inequalities

  1. #16
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    Quote Originally Posted by Jhevon View Post
    i always use \geq, that works as well.
    Gotcha! Okay, I've been copying/pasting everybody's code and modifying. Thanks for the link to the tutorial - you're good!
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  2. #17
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    This note is addressed to lilrhino
    You asked about understanding the solution to inequalities not TeX.
    As I wrote to you, it is all about the factors.
    Consider (x+a). That factor is positive for x>-a and is negative for x<-a. For (x+b) the factor is positive for x>-b and is negative for x<-b.
    Now consider (x+a)(x+b), that product is positive if both factors have the same ‘sign’; that product is negative if the factors have different ‘signs’.

    This idea can be extended to any number of factors or quotients.
    If you want to understand how inequalities work, then you need to understand the fundamental way factors contribute to the ‘signs’ of products and quotients.
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  3. #18
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    Quote Originally Posted by Plato View Post
    This note is addressed to lilrhino
    You asked about understanding the solution to inequalities not TeX.
    As I wrote to you, it is all about the factors.
    Consider (x+a). That factor is positive for x>-a and is negative for x<-a. For (x+b) the factor is positive for x>-b and is negative for x<-b.
    Now consider (x+a)(x+b), that product is positive if both factors have the same ‘sign’; that product is negative if the factors have different ‘signs’.

    This idea can be extended to any number of factors or quotients.
    If you want to understand how inequalities work, then you need to understand the fundamental way factors contribute to the ‘signs’ of products and quotients.
    Plato you are correct. Anyway, I was reading through the text and since our sign is < we only care about numbers that are negative is that correct?
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  4. #19
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    Quote Originally Posted by lilrhino View Post
    Plato you are correct. Anyway, I was reading through the text and since our sign is < we only care about numbers that are negative is that correct?
    we only care about factors that are negative
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  5. #20
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    Quote Originally Posted by Plato View Post
    we only care about factors that are negative
    Got it! I'm getting there...
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  6. #21
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    This is an inequality that's just one variable (not sure how to proceed):

    \frac{x+1}{2x-3} ≥ 0
    In my 2nd question, I came up with 3 intervals to test:

    (-\infty{, -1})
    (-1, \frac{3}{2})
    (\frac{3}{2}, \infty)

    But after testing (-4, 0, 3) I found that the factors were only negative in:

    (-\infty{, -1})
    (-1, \frac{3}{2})

    So, is this my correct solution set?
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  7. #22
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    Quote Originally Posted by lilrhino View Post
    In my 2nd question, I came up with 3 intervals to test:

    (-\infty{, -1})
    (-1, \frac{3}{2})
    (\frac{3}{2}, \infty)

    But after testing (-4, 0, 3) I found that the factors were only negative in:

    (-\infty{, -1})
    (-1, \frac{3}{2})

    So, is this my correct solution set?
    Your intervals were correct, but your interpretation was wrong. All the factors do is tell you where to split the intervals.

    Look at the actual problem:
    \frac{x + 1}{2x - 3} \geq 0

    We wish to test if \frac{x + 1}{2x - 3} is positive, not just the factors.

    So test the expression on each interval:
     ( -\infty, -1] \to \frac{x + 1}{2x - 3} \geq 0 (Check!)
     \left ( -1, \frac{3}{2} \right ) \to \frac{x + 1}{2x - 3} < 0 (No!)
     \left ( \frac{3}{2}, \infty \right ) \to \frac{x + 1}{2x - 3} > 0 (Check!)

    So the solution set is ( -\infty, -1] \cup \left ( \frac{3}{2}, \infty \right ) . (Note: we need to include x = -1, but can't include x = 3/2. Do you see why?)

    -Dan
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  8. #23
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    Quote Originally Posted by topsquark View Post
    Your intervals were correct, but your interpretation was wrong. All the factors do is tell you where to split the intervals.

    Look at the actual problem:
    \frac{x + 1}{2x - 3} \geq 0

    We wish to test if \frac{x + 1}{2x - 3} is positive, not just the factors.

    So test the expression on each interval:
     ( -\infty, -1] \to \frac{x + 1}{2x - 3} \geq 0 (Check!)
     \left ( -1, \frac{3}{2} \right ) \to \frac{x + 1}{2x - 3} < 0 (No!)
     \left ( \frac{3}{2}, \infty \right ) \to \frac{x + 1}{2x - 3} > 0 (Check!)

    So the solution set is ( -\infty, -1] \cup \left ( \frac{3}{2}, \infty \right ) . (Note: we need to include x = -1, but can't include x = 3/2. Do you see why?)

    -Dan
    Is it because it will cause the denominator to equal 0?
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  9. #24
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by lilrhino View Post
    Is it because it will cause the denominator to equal 0?
    Yup!

    -Dan
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  10. #25
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by topsquark View Post
    Your intervals were correct, but your interpretation was wrong. All the factors do is tell you where to split the intervals.

    Look at the actual problem:
    \frac{x + 1}{2x - 3} \geq 0

    We wish to test if \frac{x + 1}{2x - 3} is positive, not just the factors.

    So test the expression on each interval:
     ( -\infty, -1] \to \frac{x + 1}{2x - 3} \geq 0 (Check!)
     \left ( -1, \frac{3}{2} \right ) \to \frac{x + 1}{2x - 3} < 0 (No!)
     \left ( \frac{3}{2}, \infty \right ) \to \frac{x + 1}{2x - 3} > 0 (Check!)

    So the solution set is ( -\infty, -1] \cup \left ( \frac{3}{2}, \infty \right ) . (Note: we need to include x = -1, but can't include x = 3/2. Do you see why?)

    -Dan
    As I did with Plato's solution, I'd like to illustrate that Dan's solution is correct with a graph. See below
    Attached Thumbnails Attached Thumbnails Help with solving inequalities-rationalgraph.jpg  
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  11. #26
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    Quote Originally Posted by topsquark View Post
    Yup!

    -Dan
    Thanks Dan, I appreciate your help and the explanation !
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  12. #27
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    Quote Originally Posted by Jhevon View Post
    As I did with Plato's solution, I'd like to illustrate that Dan's solution is correct with a graph. See below
    Thanks as always Jhevon. Do you mind sharing how you're generating the graphs .
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  13. #28
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by lilrhino View Post
    Thanks as always Jhevon. Do you mind sharing how you're generating the graphs .
    i used a program called graph which TPH introduced to me
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  14. #29
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    Quote Originally Posted by lilrhino View Post
    Please help with solving the following inequalities (2):
    x^2 + 4x - 32 < 0
    Here is another way.

    x^2 + 4x - 32 = x^2 + 4x +4 - 36 = (x+2)^2 - 36

    Thus,
    (x+2)^2 - 36 \geq 0 \Rightarrow (x+2)^2 \geq 36

    Thus,
    x+2 \geq 6 \mbox{ or }x+2 \leq -6

    The rest is trivial.
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  15. #30
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    Quote Originally Posted by lilrhino View Post
    \frac{x+1}{2x-3} \geq 0
    Here is another way.

    Either the numerator is non-negative and the denominator is postive.

    Or,

    EIther the denoimator is non-positive and the denoimator is negative.

    Thus,
    x+1 \geq 0 \mbox{ and } 2x - 3>0
    Or,
    x+1\leq 0 \mbox{ and }2x- 3<0

    The rest is trivial.
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