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    Hi I would like to know how would you be able to find out the number of solutions to the equation without working it out? For example this question. Name:  1459799107832-1339212355.jpg
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Size:  323.5 KBAttachment 519301519303 I am referring to question 10. It doesn't ask for any working just simply state the number of solutions. However my initial guess would be that there will are 2. Thanks.
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    (Original post by coconut64)
    Hi I would like to know how would you be able to find out the number of solutions to the equation without working it out? For example this question. Name:  1459799107832-1339212355.jpg
Views: 58
Size:  323.5 KBAttachment 519301519303 I am referring to question 10. It doesn't ask for any working just simply state the number of solutions. However my initial guess would be that there will be 2. Thanks.
    Can you see why the number of solutions will be the same thing as the number of times the two curves you've sketched intersect one another?

    If not, equate the two curves, expand and simplify, what do you get?
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    Also, I am wondering does the graph has to be really precise. Because sometimes I am not certain whether the turning point would be in the negative x axis or y axis. Since the question does not require me to use calculus... I have looked at the mark scheme, sometimes it doesnt show which side the turning point is specifically... So does it matter if I get the turning point in the wrong aide of the graph? Thanks!
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    (Original post by coconut64)
    Also, I am wondering does the graph has to be really precise. Because sometimes I am not certain whether the turning point would be in the negative x axis or y axis. Since the question does not require me to use calculus... I have looked at the mark scheme, sometimes it doesnt show which side the turning point is specifically... So does it matter if I get the turning point in the wrong aide of the graph? Thanks!
    In the actual exam you will be expected to place the turning point in the correct quadrant at the very least and label it with co-ordinates. The questions are unlike these ones though so when doing these ones, I don't think it'd matter too much.
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    (Original post by Zacken)
    Can you see why the number of solutions will be the same thing as the number of times the two curves you've sketched intersect one another?

    If not, equate the two curves, expand and simplify, what do you get?
    I am not quite sure what you mean here. I get that point of intersection is when two graphs cross at the same point. Of they cross once there will be one solution. However my graph will not be accurate as it's only a rough drawing. I wouldn't really know if they do cross in the negative axis unless I work it out. However the question doesnt require me to, so I thought there will be a faster way to work this out maybe?
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    (Original post by Zacken)
    In the actual exam you will be expected to place the turning point in the correct quadrant at the very least and label it with co-ordinates. The questions are unlike these ones though so when doing these ones, I don't think it'd matter too much.
    Why will the exam questions be different to these ones? I thought they are similar..
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    (Original post by coconut64)
    Why will the exam questions be different to these ones? I thought they are similar..
    They're not. You'll see once you start doing past papers.
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    (Original post by Zacken)
    They're not. You'll see once you start doing past papers.
    True. I guess the one in the exam papers are easier. Can u answer my question above please. Thanks.
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    (Original post by coconut64)
    I am not quite sure what you mean here. I get that point of intersection is when two graphs cross at the same point. Of they cross once there will be one solution. However my graph will not be accurate as it's only a rough drawing. I wouldn't really know if they do cross in the negative axis unless I work it out. However the question does require me to, so I thought there will be a faster way to work this out maybe?
    Basically the question is asking you to, from your rough sketch, state how many points of intersection there are. That's it. Use some logic to determine whether the two curves cross in the third quadrant or not, although you should have thought about that prior to drawing the sketch. Why don't you test some points. Is the first curve higher than the second for x = -1000? What about for x = -1? If it is still higher then there weren't any intersections, if one curves became higher then there must have been an intersection, etc...
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    (Original post by Zacken)
    Basically the question is asking you to, from your rough sketch, state how many points of intersection there are. That's it. Use some logic to determine whether the two curves cross in the third quadrant or not, although you should have thought about that prior to drawing the sketch. Why don't you test some points. Is the first curve higher than the second for x = -1000? What about for x = -1? If it is still higher then there weren't any intersections, if one curves became higher then there must have been an intersection, etc...
    Oh right. I guess that is same for turning points then. Thank you for helping again!
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    (Original post by coconut64)
    Oh right. I guess that is same for turning points then. Thank you for helping again!
    No problem! Sorry to not be of more use. :-(
 
 
 
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