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    I've managed to do the first part on question 6 but I'm stuck on part b.
    Here's the question:

    Find the value or set of values of k such that the equation x^2 + 2x + k = 0 has real and distinct roots.
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    Here's my working out so far: Name:  263AD1F9-9590-469C-8F7E-5C7560DE46CA.jpg.jpeg
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    dont we need the first part to help

    (Original post by Emsio8)
    I've managed to do the first part on question 6 but I'm stuck on part b.
    Here's the question:

    Find the value or set of values of k such that the equation x^2 + 2x + k = 0 has real and distinct roots.
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    (Original post by angrypoliceman)
    dont we need the first part to help
    Okay. Here's the first part and my working out: Name:  F195729F-A92C-43DE-BC32-6BD4AC9121DB.jpg.jpeg
Views: 22
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    Name:  7135EC98-1C7C-41E8-A41F-1F5597385E0F.jpg.jpeg
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    (Original post by Emsio8)
    Okay. Here's the first part and my working out:
    uhhhh **** that dude put it the right way round
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    (Original post by angrypoliceman)
    uhhhh **** that dude put it the right way round
    You aren't being very helpful. Why can't you see it that way around?
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    (Original post by Emsio8)
    Here's my working out so far:
    c=k here, and when dividing by a negative number, the inequality flips, AND the RHS doesn't go to 0...
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    (Original post by Emsio8)
    You aren't being very helpful. Why can't you see it that way around?
    im not straining my neck to help ur stuck ass, still love u tho cause ur human
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    (Original post by RDKGames)
    c=k here, and when dividing by a negative number, the inequality flips.
    But that would mean the equation would no longer have two real and distinct roots. It would have no real roots.

    It said in the question two real and distinct roots?
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    (Original post by Emsio8)
    But that would mean the equation would no longer have two real and distinct roots. It would have no real roots.

    It said in the question two real and distinct roots?
    We can find the two distinct roots later but the condition here for real and distinct roots are (as you stated):

     b^{2} - 4ac > 0.

    So you just substitute the coefficients a, b and c into the equation.

    --------------------------------------------------------------------------------------------------------------

    For your second post of this thread, as RDKGames has pointed out:
    -Dividing by a negative value usually reverses the inequality sign, try adding/subtracting terms instead.

    -Fourth line: You have dropped the inequality condition.

    -Fifth line: You need to look at your working as:

     (-4)(0) \neq -4.
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    (Original post by simon0)
    We can find the two distinct roots later but the condition here for real and distinct roots are (as you stated):

     b^{2} - 4ac > 0.

    So you just substitute the coefficients a, b and c into the equation.

    Also as RDKGames has pointed out (for your second post of this thread):
    -Dividing by a negative value reverses the sign of the inequality, try subtracting the term instead.

    -You need to look at the fifth line of your working:

     (-4)(0) \neq -4.
    Is this looking better? If so, where do I go from here?Name:  967E3C73-3592-4E3B-85F3-EA457099D2BB.jpg.jpeg
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    (Original post by Emsio8)
    Is this looking better? If so, where do I go from here?
    You've done it. Just y'know, replace c with k for obvious reasons.
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    (Original post by RDKGames)
    You've done it. Just y'know, replace c with k for obvious reasons.
    Thanks for the help!
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    (Original post by RDKGames)
    c=k here, and when dividing by a negative number, the inequality flips, AND the RHS doesn't go to 0...
    is there a way for the images to be rotated ? it often happens that they are sideways & are not easy to read :dontknow:
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    (Original post by the bear)
    is there a way for the images to be rotated ? it often happens that they are sideways & are not easy to read :dontknow:
    I understand, but there isn't a way atm unless we somehow prod those TSR techies in their towers. I just use a chrome extension to rotate the pages and actually read the images.
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    (Original post by RDKGames)
    I understand, but there isn't a way atm unless we somehow prod those TSR techies in their towers. I just use a chrome extension to rotate the pages and actually read the images.
    i just installed it too & problem solved ! ( it is IMG Rotate )

    thanks for the tip !
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    (Original post by the bear)
    i just installed it too & problem solved ! ( it is IMG Rotate )

    thanks for the tip !
    Ooh that's a nice extension there - and didn't appear anywhere on the extension store for me at first! I was using a different one which had some practical inconveniences as it rotated whole pages (so I opened images in a new tab) but this one does the job exactly as I want it, thanks
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    (Original post by RDKGames)
    Ooh that's a nice extension there - and didn't appear anywhere on the extension store for me at first! I was using a different one which had some practical inconveniences as it rotated whole pages (so I opened images in a new tab) but this one does the job exactly as I want it, thanks
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    (Original post by RDKGames)
    c=k here, and when dividing by a negative number, the inequality flips, AND the RHS doesn't go to 0...
    It's VERY bad practice to suggest this - far better to suggest collecting so the coefficient of the unknown is positive.
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    (Original post by simon0)

    For your second post of this thread, as RDKGames has pointed out:
    -Dividing by a negative value reverses the inequality sign.
    PLEASE don't ever suggest this approach ... it is always possible to avoid and people remember it incorrectly or don't understand why.
 
 
 
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