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    Hi all, just a quick question about a C1 maths past paper question. I'm not entirely sure how to go about answering this question, anyway here it is.
    Find the set of values of  k for which the graph of  y=x^2 + 2kx + 5 does not intersect the  x axis.
    Any help on how to answer this question or the method I should use will be much appreciated.
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    When a parabola does not intersect the x axis, the discriminant of the quadratic will be negative.
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    (Original post by biologeek)
    Hi all, just a quick question about a C1 maths past paper question. I'm not entirely sure how to go about answering this question, anyway here it is.
    Find the set of values of  k for which the graph of  y=x^2 + 2kx + 5 does not intersect the  x axis.
    Any help on how to answer this question or the method I should use will be much appreciated.
    I believe this means that the curve does not cross the x-axis, so therefore
     b^2-4ac<0
    Input the values which give us
     4k^2-20<0
    Which means
     4k^2<20

k^2<5

k is less than \sqrt5

    Was this the right answer?
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    (Original post by MashB)
    I believe this means that the curve does not cross the x-axis, so therefore
     b^2-4ac<0
    Input the values which give us
     4k^2-20<0
    Which means
     4k^2<20

k^2<5

k is less than \sqrt5

    Was this the right answer?
    No. That isn't how you solve quadratic inequalities.
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    (Original post by Mr M)
    No. That isn't how you solve quadratic inequalities.
    How do you?


    This was posted from The Student Room's iPhone/iPad App
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    (Original post by MashB)
    How do you?
    The easiest way is to graph the inequality.

    k^2 <5 means k^2-5<0

    Sketch y=k^2-5

    Check which values of k produce negative values of y.
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    (Original post by Mr M)
    The easiest way is to graph the inequality.

    k^2 <5 means k^2-5<0

    Sketch y=k^2-5

    Check which values of k produce negative values of y.
    (Just checking)  -\sqrt{5} < k < \sqrt{5} ?
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    (Original post by McLH)
    (Just checking)  -\sqrt{5} < k < \sqrt{5} ?
    Yes.
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    (Original post by Mr M)
    Yes.
    So in a way I was on the right track, just didnt finish it...
     k<\pm\sqrt5
    k=\sqrt5 and  k=-\sqrt5
    so plot them on the graph and then that becomes
    -\sqrt5<k<\sqrt5
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    (Original post by MashB)
    So in a way I was on the right track, just didnt finish it...
     k<\pm\sqrt5
    That does not lead to this

    -\sqrt5<k<\sqrt5
    The first line is not possible
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    (Original post by TenOfThem)
    That does not lead to this



    The first line is not possible
    Could you explain it, I'm confusing myself here now.


    This was posted from The Student Room's iPhone/iPad App
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    (Original post by MashB)
    Could you explain it, I'm confusing myself here now.
    You go from

    k^2<5

    to

    -\sqrt{5}<k<\sqrt{5}

    there are no intermediate lines
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    (Original post by TenOfThem)
    You go from

    k^2<5

    to

    -\sqrt{5}<k<\sqrt{5}

    there are no intermediate lines
    I meant as in explain the answer to op's question.


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    (Original post by MashB)
    I meant as in explain the answer to op's question.
    the answer is in the thread

    not sure what else you want
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    (Original post by MashB)
    Could you explain it, I'm confusing myself here now.


    This was posted from The Student Room's iPhone/iPad App
    I know I shouldn't post full solutions, but for benefit:

     b^2 -4ac < 0

     (2k)^2 - (4 * 1 * 5) < 0

     4k^2 - 20 < 0

    Divide by 4
     k^2 - 5 < 0

     k^2 < 5

     k = \pm\sqrt{5}

    Plot
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    (Original post by McLH)


     k = \pm\sqrt{5}
    This line is incorrect
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    (Original post by TenOfThem)
    This line is incorrect
    As a solution, but to show where the roots are that I'm plotting before i do a sketch of the graph is it? If so, what should I write?
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    (Original post by McLH)
    As a solution, but to show where the roots are that I'm plotting before i do a sketch of the graph is it? If so, what should I write?
    The answer that I gave in post 12
 
 
 
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