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    I'm doing C3 Jan 2012 on AQA and currently on Q7a. I got the wrong answer but my method is the same (I have what's on line 2) up until it says that the expression ≠ 0 (line 3, second image)
    Name:  Screen Shot 2013-04-05 at 10.38.28.png
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    Name:  Screen Shot 2013-04-05 at 10.38.40.png
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    I don't understand how that helps to figure out the answer?
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    (Original post by PrinceyJ)
    I'm doing C3 Jan 2012 on AQA and currently on Q7a. I got the wrong answer but my method is the same (I have what's on line 2) up until it says that the expression ≠ 0 (line 3, second image)
    Name:  Screen Shot 2013-04-05 at 10.38.28.png
Views: 78
Size:  23.2 KB

    Name:  Screen Shot 2013-04-05 at 10.38.40.png
Views: 91
Size:  53.6 KB
    I don't understand how that helps to figure out the answer?
    Note that the graph of e to the anything is always above the x axis, so it cannot be 0 (so the other bracket must be)
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    (Original post by PrinceyJ)
    I'm doing C3 Jan 2012 on AQA and currently on Q7a. I got the wrong answer but my method is the same (I have what's on line 2) up until it says that the expression ≠ 0 (line 3, second image)
    Name:  Screen Shot 2013-04-05 at 10.38.28.png
Views: 78
Size:  23.2 KB

    Name:  Screen Shot 2013-04-05 at 10.38.40.png
Views: 91
Size:  53.6 KB
    I don't understand how that helps to figure out the answer?
    a \times b = 0 can be satisfied if a = 0 or b = 0 or both.

    In this case, e^y is greater than zero for all real y so you only get the two solutions from the quadratic. The fact that it does not equal zero does not help to solve the quadratic; it just tells you that you don't need to look for any more solutions.

    Note that if you factor out x from the quadratic, you check that x = 0 works and setting the remaining linear expression equals zero works, too.
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    (Original post by Indeterminate)
    Note that the graph of e to the anything is always above the x axis, so it cannot be 0 (so the other bracket must be)
    (Original post by Brister)
    a \times b = 0 can be satisfied if a = 0 or b = 0 or both.

    In this case, e^y is greater than zero for all real y so you only get the two solutions from the quadratic. The fact that it does not equal zero does not help to solve the quadratic; it just tells you that you don't need to look for any more solutions.

    Note that if you factor out x from the quadratic, you check that x = 0 works and setting the remaining linear expression equals zero works, too.
    Thanks guys, I wasn't really in the correct frame of thinking when I came across this
 
 
 
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