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    Did the question ask for the range of values where the equation has roots, or where it doesn't?
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    (Original post by UCASLord)
    I'm a bit confused about why the answer to this inequality turns out as it does:

    Find the range of values for k for which the equation x^2 - kx + (k + 3) = 0

    My working:

    b^2 - 4ac > 0
    k^2 - 4k - 12 > 0
    (k - 6)(k + 2) > 0
    so critical points are at x = -2 and x = 6

    I draw the graph, and see when the graph of this quadratic is above 0.

    Which is when k < -2 or x > 6.

    But the answer is -2 < k < 6?
    check that you've read the question properly. That interval contains the values of k for which there are no real roots.
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    (Original post by UCASLord)
    I know - that's why I created this thread.

    I've definitely read this question properly.

    It's question 4a) in the Edexcel C1 textbook, section 3.

    At first, I thought it was a textbook error, but question 4b) also gives the interval for which the values of k have no read roots, which is very confusing.
    you mean they gave the same interval for a) and b)? Then I'd suggest an error in the textbook.
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    Note that it's obvious the equation has no roots when k=0.
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    (Original post by UCASLord)
    Hey Franklin, missed your reply at first.

    It's question 4a) from the Edexcel C1 textbook, section 3. I'm sure I read the question correctly.
    I think I have a digital copy of this actually, I'll have a look. What page is it exactly?
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    (Original post by UCASLord)
    Well spotted, because then 3 = 0, which is not true, and yet 0 is within the range of solutions given in the answer section.

    I guess the writer must have been a bit tired on the day :P
    Well, not quite, you'd have x^2 + 3 = 0 , which has no real solutions.
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    (Original post by UCASLord)
    I'm a bit confused about why the answer to this inequality turns out as it does:

    Find the range of values for k for which the equation x^2 - kx + (k + 3) = 0

    My working:

    b^2 - 4ac > 0
    k^2 - 4k - 12 > 0
    (k - 6)(k + 2) > 0
    so critical points are at x = -2 and x = 6

    I draw the graph, and see when the graph of this quadratic is above 0.

    Which is when k < -2 or x > 6.

    But the answer is -2 < k < 6?
    It's a typo that I think might have been corrected in later editions of the textbook. Here's the correct version that is in the Solutionbank:

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