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    i understand the topic (c1) but i always get the answer wrong (the last part where you have to write for e.g 3>x>-4 if the graph is for e.g x>0
    i dont really understand?
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    (Original post by seconda1)
    i understand the topic (c1) but i always get the answer wrong (the last part where you have to write for e.g 3>x>-4 if the graph is for e.g x>0
    i dont really understand?
    Try this video.

    If there's a specific part of the method that you don't understand then please let us know.
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    (Original post by Notnek)
    Try this video.

    If there's a specific part of the method that you don't understand then please let us know.
    he x(-1) but i thought you were supposed to still draw the original -x^2 graph? theres an example of this in the textbook
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    (Original post by seconda1)
    he x(-1) but i thought you were supposed to still draw the original -x^2 graph? theres an example of this in the textbook
    Your question is unclear. Could you please explain in more detail and tell us the part of the video (give the time) that you don’t understand?
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    (Original post by Notnek)
    Your question is unclear. Could you please explain in more detail and tell us the part of the video (give the time) that you don’t understand?
    okay i dont understand the last part when finding where y is > or < than 0 i understand how to factorise and draw the graph/find critical values.

    so when y > 0 i look above the axis but i dont understand how to get the value?

    thanks
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    (Original post by seconda1)
    okay i dont understand the last part when finding where y is > or < than 0 i understand how to factorise and draw the graph/find critical values.

    so when y > 0 i look above the axis but i dont understand how to get the value?

    thanks
    Let's say you were solving x^2+x-2&gt;0

    Factorise to get

    (x-1)(x+2)&gt;0

    The critical points (roots) are 1 and -2.


    Then sketch the graph of y=x^2+x-2:




    We need x^2+x-2&gt;0 which means we are looking for where y&gt;0. This occurs above the x-axis and this is shown above as the lines highlighted in red. If you now consider the x values for this red region, you can see that x&gt;1 or x&lt;-2 so this is the solution.

    If instead you were solving x^2+x-2&lt;0 then you need to find where y&lt;0. This occurs below the x-axis which is the blue region on the graph above. This region occurs where -2&lt;x&lt;1 so this is the solution.


    If any of this doesn't make sense please let me know the specific part that confuses you.
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    (Original post by Notnek)
    Let's say you were solving x^2+x-2&gt;0

    Factorise to get

    (x-1)(x+2)&gt;0

    The critical points (roots) are 1 and -2.


    Then sketch the graph of y=x^2+x-2:




    We need x^2+x-2&gt;0 which means we are looking for where y&gt;0. This occurs above the x-axis and this is shown above as the lines highlighted in red. If you know consider the x values for this red region, you can see that x&gt;1 or x&lt;-2 so this is the solution.

    If instead you were solving x^2+x-2&lt;0 then you need to find where y&lt;0. This occurs below the x-axis which is the blueregion on the graph above. This region occurs where -2&lt;x&lt;1 so this is the solution.


    If any of this doesn't make sense please let me know the specific part that confuses you.

    thanks soo much. i think i finally understand. in videos they dont go through it clear enough for me because i take time to understand.
    i think i will highlight where y > or < 0

    in the case of it being equal there are two separate solutions right?
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    (Original post by seconda1)
    in the case of it being equal there are two separate solutions right?
    I'm not sure what you mean here.
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    (Original post by Notnek)
    I'm not sure what you mean here.
    when the question asks where y is greater or equal to 0
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    (Original post by seconda1)
    when the question asks where y is greater or equal to 0
    If the question was to solve x^2+x-2\geq 0 instead of x^2+x-2&gt;0 then all you do is change the solutions from &gt; or &lt; to \geq or \leq.

    So x&gt;1 or x&lt;-2 becomes

    x\geq 1 or x\leq -2.
 
 
 
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