Find the range of values of k for which the equation x^2 - kx + (k+3) = 0 has no real

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    Find the range of values of k for which the equation x^2 - kx + (k+3) = 0 has no real roots.

    Could I have some help with this question please?

    I have got this far:

    b^2 - 4ac < 0

    k^2 - (4 x 1 x (k+3)) < 0

    k^2 - 4k +12 = 0

    (k - 6) (k + 2) = 0

    k = 6 or k = -2

    Afterwards I attempted to draw a graph of k^2 - 4k +12 = 0, however it does not cross the x-axis? Firstly, how is it possible to find solutions to an equation that does not cross the x-axis?

    I was going to shade in the areas on the below the x-axis, as this would've given me values for where the equation is less than 0, but it doesn't cross the x-axis?

    How should I proceed? Thanks for any help!
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    For no real roots  b^2-4ac&lt;0
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    b^2 is negative rather than positive (check the original quadratic).
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    (Original post by TheOtherSide.)
    b^2 is negative rather than positive (check the original quadratic).
    Thanks for the reply

    b is -k

    b^2 is (-k)^2, which is k^2

    Am I missing something?
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    (Original post by NotNotBatman)
    For no real roots  b^2-4ac&lt;0
    Thanks for the reply,

    yeah sorry that was a silly slip when I was typing out my working, I shall edit it now, thanks for spotting it
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    (Original post by TheOtherSide.)
    b^2 is negative rather than positive (check the original quadratic).
    b is negative (-k), but b^2 is positive (-k)^2 = k^2
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    (Original post by tazarooni89)
    b is negative (-k), but b^2 is positive (-k)^2 = k^2
    That's what I meant - I probably should have been clearer..

    (Original post by machete-monster1)
    Thanks for the reply

    b is -k

    b^2 is (-k)^2, which is k^2

    Am I missing something?
    Your actual equation would be -k^2 - 4k - 12, which suggests what about the shape of the curve?
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    (Original post by machete-monster1)
    Find the range of values of k for which the equation x^2 - kx + (k+3) = 0 has no real roots.

    Could I have some help with this question please?

    I have got this far:

    b^2 - 4ac < 0

    k^2 - (4 x 1 x (k+3)) < 0

    k^2 - 4k +12 = 0

    (k - 6) (k + 2) = 0

    k = 6 or k = -2

    Afterwards I attempted to draw a graph of k^2 - 4k +12 = 0, however it does not cross the x-axis? Firstly, how is it possible to find solutions to an equation that does not cross the x-axis?

    I was going to shade in the areas on the below the x-axis, as this would've given me values for where the equation is less than 0, but it doesn't cross the x-axis?

    How should I proceed? Thanks for any help!
    check that bolded line, you have made a mistake in expanding the bracket above

    but you've somehow factorized right :laugh:
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    (Original post by TheOtherSide.)
    That's what I meant - I probably should have been clearer..



    Your actual equation would be -k^2 - 4k - 12, which suggests what about the shape of the curve?
    The equation is B^2 - 4AC < 0
    Where B = (-k)
    => k^2 - 4(k+3) < 0

    There should still be no negative sign before the k^2

    The curve is a "U" shaped parabola, not an "n" shaped one.*
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    (Original post by tazarooni89)
    The equation is B^2 - 4AC < 0
    Where B = (-k)
    => k^2 - 4(k+3) < 0

    There should still be no negative sign before the k^2

    The curve is a "U" shaped parabola, not an "n" shaped one.*
    Okay, I understand. But surely then the second equation will have no solutions?
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    (Original post by TheOtherSide.)
    Okay, I understand. But surely then the second equation will have no solutions?
    Wait, I see.
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    OP, check the y-intercept in the second equation..
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    (Original post by TheOtherSide.)
    OP, check the y-intercept in the second equation..
    Thanks of your help

    Y intercept is 12

    Is this incorrect?
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    (Original post by machete-monster1)
    Thanks of your help

    Y intercept is 12

    Is this incorrect?
    Slightly - check your bracket expansion.
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    (Original post by machete-monster1)
    Find the range of values of k for which the equation x^2 - kx + (k+3) = 0 has no real roots.

    Could I have some help with this question please?

    I have got this far:

    b^2 - 4ac < 0

    k^2 - (4 x 1 x (k+3)) < 0

    k^2 - 4k +12 = 0

    (k - 6) (k + 2) = 0

    k = 6 or k = -2

    Afterwards I attempted to draw a graph of k^2 - 4k + 12 = 0, however it does not cross the x-axis? Firstly, how is it possible to find solutions to an equation that does not cross the x-axis?

    I was going to shade in the areas on the below the x-axis, as this would've given me values for where the equation is less than 0, but it doesn't cross the x-axis?

    How should I proceed? Thanks for any help!
    You used the wrong sign here (should be k^2 - 4k - 12). Also, you should keep the inequalities in the equation, then sketch the graph to find your values of k.
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    Thank you very much everyone for the help; this was such a silly mistake!
 
 
 
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