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    Hi all,

    I need help with 5 (b) (i), please. I have tried to find discriminant (D = b^2 - 4ac) but to no avail.

    Please do explain this.

    Thank you very much!
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    (Original post by londoncricket)
    Hi all,

    I need help with 5 (b) (i), please. I have tried to find discriminant (D = b^2 - 4ac) but to no avail.

    Please do explain this.

    Thank you very much!
    As you've said we have \displaystyle (x+1)(x-3) = kx(x-2)

    Expanding gives us:

    \displaystyle x^2 -2x -3 = kx^2 - 2kx \Rightarrow x^2(k-1) + x(2 - 2k) + 3 = 0

    So what's the discriminant of this quadratic?
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    Thank you for your prompt reply!

    The discriminant of this quadratic would be:

    4k^2 - 20k + 16

    (Sorry, I am not well versed in LaTeX).
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    (Original post by londoncricket)
    Thank you for your prompt reply!

    The discriminant of this quadratic would be:

    4k^2 - 20k + 16
    Now you know that that the discriminant should be \geq 0, since the "intersect" means that there is either one or two solutions.

    This gives us 4k^2 - 20k + 16 > 0 \Rightarrow 4(k^2 - 5k + 4)> 0 \Rightarrow k^2 - 5k + 4 > 0. By dividing by 4.

    Could you, perchance - factorise this for me?

    (Sorry, I am not well versed in LaTeX).
    Don't worry one iota, you're coming across perfectly clear. You do want to press the "reply" button at the bottom-right corner of my post so that I get a notification when you reply to me. :-)
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    (Original post by Zacken)
    Now you know that that the discriminant should be \geq 0, since the "intersect" means that there is either one or two solutions.

    This gives us 4k^2 - 20k + 16 > 0 \Rightarrow 4(k^2 - 5k + 4)> 0 \Rightarrow k^2 - 5k + 4 > 0. By dividing by 4.

    Could you, perchance - factorise this for me?



    Don't worry one iota, you're coming across perfectly clear. You do want to press the "reply" button at the bottom-right corner of my post so that I get a notification when you reply to me. :-)
    That's great. Thanks very much!

    I was getting those answers before, but I did not think to factorise the values derived for k, hence not getting it in the form the question wanted it in.

    Thanks a tonne for your help! Definitely deserve a rep for this.
 
 
 
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