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Thread: Solving an inequality

  1. #1 Solving an inequality 
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    I'm looking at the following question:

    Solve the inequality y^2 + 2y is greater than or equal to 0.

    I guess that the first step is to factorize and get y(y+2) is greater than or equal to 0, but after that I'm not sure what needs to be done.

    Any help would be great.


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  3. #2  
    Forum Ph.D. Heinsbergrelatz's Avatar
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    ,
    ,


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  4. #3  
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    Thanks.

    So, what we've done there is find out where on the curve the line is above (on the positive side of) the x-axis i.e. everywhere apart from the values x=-2 through to x=0. Is that correct?

    Also, can you talk me through the step from: y(y+2)is greater than or equal to 0,
    to: y=greater than or equal to 0, y=greater than or equal to -2?
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  5. #4  
    Forum Ph.D. Heinsbergrelatz's Avatar
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    Yes we choose the interval where the curve is above the x-axis. But there are two common conditions;


    ,


    so in the question above, what happened was it reached to the point ;









    thus the inequality sign changes;

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  6. #5  
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    Cheers! That's much clearer. I really appreciate your taking the time to help.
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  7. #6  
    Forum Professor jrmonroe's Avatar
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    Before wracking our brains with the math, let’s start by mentally inspecting the original expression.

    It is a relatively simple expression. The second-order term (ie, the term) will produce a parabola, which generally means that the solution either includes or excludes a bounded region (as we shall see). The positive sign of this second-order term shows that it is a parabola that “opens”, or extends, into the positive direction. The inequality is “greater than...”. Together, these two facts (the positive sign and the "greater than" inequality) mean generally that the solution excludes a bounded region (ie, the solution extends beyond both sides of the bounded region ... the bounded region being the bottom part of the parabola that dips below zero).

    Even without the inspection, actually solving this expression can begin here.

    Replace the inequality function with an equality function (ie, an equal sign), then solve to provide the values at which the expression intersects the zero line. These intersections are part of the solution because the original operator is “greater than or equal to”. Then test both sides of these intersections to determine the regions that fall within the expression’s criteria. This provides the answer.

    Regardless of the method for arriving at a solution, it always pays to test the solution by plugging values back into the original expression. Both teachers and bosses expect this. Using a spreadsheet and using columns of , , and values with small enough increments for will show exactly why this expression behaves as it does.
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  8. #7  
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    Actually, I've had another look, and the values of y will actually be y>=0 and y<=-2.

    For example, if we plug in -1, then we have: 1-2>=0, which is not true, so this has to be excluded.

    But I can't figure out at what point the sign changes to yield y<=-2.
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  9. #8  
    Forum Ph.D. Heinsbergrelatz's Avatar
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    Actually, I've had another look, and the values of y will actually be y>=0 and y<=-2.

    For example, if we plug in -1, then we have: 1-2>=0, which is not true, so this has to be excluded.

    But I can't figure out at what point the sign changes to yield y<=-2.
    that is what i thought, but then the signs doesn't make much sense. Of course with the solutions i gave, it violates the inequality with -1.

    im not quite sure if its
    or

    let Dr. Rocket or Arcane for the confirmation
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  10. #9  
    Forum Professor jrmonroe's Avatar
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    Using a spreadsheet and using columns of , , and values with small enough increments for (let's say increments of 0.25 over the range from –3 to +1) will show exactly why this expression behaves as it does.

    About the signs ...

    When < –2, both and are negative, which give a positive when multiplied together.

    When > 0, both and are positive, which give a positive when multiplied together.

    When –2 < < 0, then is negative and is positive, which give a negative when multiplied together.
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  11. #10  
    . DrRocket's Avatar
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    Quote Originally Posted by Heinsbergrelatz
    Actually, I've had another look, and the values of y will actually be y>=0 and y<=-2.

    For example, if we plug in -1, then we have: 1-2>=0, which is not true, so this has to be excluded.

    But I can't figure out at what point the sign changes to yield y<=-2.
    that is what i thought, but then the signs doesn't make much sense. Of course with the solutions i gave, it violates the inequality with -1.

    im not quite sure if its
    or

    let Dr. Rocket or Arcane for the confirmation
    It is or

    What you have is a product of two numbers and
    For the product to be non-negative either both factors must be non-negative or both factors must be non-positive. Both are non-negative if and both are non-positive if
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