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Find the set of values of x for which C1 Watch

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    So the answer is in fact k\le2, k\ge6
    GeoGebra agrees with me.
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    (Original post by MathMeister)
    So the answer is in fact x\le2, x\ge6
    You mean k, x doesn't come into it.
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    (Original post by MathMeister)
    So, to conclude this question, a real root is when there are 2 x-intercepts. A real root cannot be an equal root (vice versa) as an equal root does not cross the x-axis. Am I correct in thinking this?
    Also, (sorry) another question I have came up with, is why the graphical representation of the expression tells you what values of k for which the equation has real roots. I see the inequality, but fail to see how this relates to k. I don't see.... ahhh wait... I figured it out. This has been the best maths problem so far. Wopee! I completely and utterly understand it all now. I understand the question and answer entirely. Please may somebody answer the underlined question however, still.
    Thank you!
    Sorry about the confusion here. A real root can be an equal root; when this happens, the graph touches the x-axis, at one point, rather than crossing it at two different points.

    The reason why I didn't tell you to put the greater than or equal to sign in, is that when  b^2 - 4ac equals zero, there are equal roots, which with the inequalities, rarely happens.
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    (Original post by Khallil)
    Remember that both inequalities have to be satisfied.

    \begin{aligned} x^2 - x - 6 > 0 & \implies x^2 - 3x + 2x - 6 > 0 \\ & \implies (x+2)(x-3) > 0 \\ & \implies x \in \left( -\infty, -2 \right) \cup \left( 3, +\infty \right) \end{aligned}

    10 - 2x < 5 \implies 5 < 2x \implies 2.5 < x

    These are both satisfied by x > 3

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    If I wanted to use mathematical notation to show the set of values of x for which both inequalities are satisfied, would this be ok?

     \left( \left( -\infty, -2 \right) \cup \left( 3, +\infty \right) \right) \cap \left( 2.5, +\infty \right) = \left( 3, + \infty)
    I'm sorry Khalil but this maths is gibberish to me. I don't get the infinity thing. :/
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    (Original post by MathMeister)
    I'm sorry Khalil but this maths is gibberish to me. I don't get the infinity thing. :/
    That means a set of numbers

    The closed interval (-infty, 2] means that x can take any value between minus infinity and two (counting two because it's closed on the right side).

    It's basically just another way of saying k <= 2
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    (Original post by spotify95)
    Sorry about the confusion here. A real root can be an equal root; when this happens, the graph touches the x-axis, at one point, rather than crossing it at two different points.

    The reason why I didn't tell you to put the greater than or equal to sign in, is that when  b^2 - 4ac equals zero, there are equal roots, which with the inequalities, rarely happens.
    What GeoGebra gives is this... And yes... arithmeticae, I meant k, not x. I suppose it does happen in this case?
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    (Original post by MathMeister)
    Prsom* ... to all-In this case then surely spotify95 was wrong in not putting greater than or equal sign in his working. If that's the case then maybe he did not know this. Somebody help?
    Also, I will try to understand the delta thing arithmetic posted (ty)
    Apologies for this, MathMeister.

    What I meant, was that the answer was going to have distinct, real roots, and not equal, real roots. It is possible to have equal, real roots - however, in the inequality we set up in that example (with the discriminant) I assumed distinct, real roots. Think of it this way.
    If  b^2 - 4ac \ge 0 , then the equation for k could have equal roots as well as distinct roots. The equal roots thing very rarely happens with inequalities.

    However, now I do agree that the answer should have been greater than or equal to, rather than just greater than etc. So apologies if I confused anyone!
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    (Original post by MathMeister)
    What GeoGebra gives is this... And yes... arithmeticae, I meant k, not x. I suppose it does happen in this case?
    Yes, I think that's correct.

    In this case it was greater than or equal to, as it didn't matter whether the roots were distinct or equal.
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    (Original post by Arithmeticae)
    ...
    My hero, qtsie tootsie :holmes:
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    (Original post by Khallil)
    My hero, qtsie tootsie :holmes:
    Has Zen hacked into your account or something? :lol:
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    (Original post by spotify95)
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    Hi,

    I'll give it a crack now. It is asking for the set of values for k, that let  b^2 - 4ac &gt; 0. In other words, what values of k can be put into the equation, that let the discriminant be more than zero?

    Here, we need to set up an inequality; that is:  b^2 - 4ac &gt; 0 .

    By plugging in the values for a, b and c, and simplifying, we get:  k^2 - 4k - 12 &gt; 0

    Solving gives the roots of the quadratic as  k = -2 or  k = 6 .

    Then, graph the curve. Remember, since the discriminant must be greater than zero for the equation to have real roots, the quadratic involving k must be greater than zero.

    Hence, you need the part of the curve that is greater than zero, which gives the answer:

     k &lt; -2 and  k &gt; 6

    I've also attached my methodology to this problem.
    Hope I've helped
    Hi there, just so you know for next time, we're not allowed to post full solutions as this doesn't help the user out, instead we provide hints and guide the user through the question to make sure they know how to do it on their own.

    I understand this time you've helped out the user by continuing to provide assistance, but please don't post full solutions again otherwise I might have to give you a warning, I hope you understand why.
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    (Original post by Arithmeticae)
    ...
    Nah, we all speak like this now (Z, K and I). :lol:
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    (Original post by MathsNerd1)
    Hi there, just so you know for next time, we're not allowed to post full solutions as this doesn't help the user out, instead we provide hints and guide the user through the question to make sure they know how to do it on their own.

    I understand this time you've helped out the user by continuing to provide assistance, but please don't post full solutions again otherwise I might have to give you a warning, I hope you understand why.
    Okay thanks - I was wondering whether posting full solutions was allowed on here or not (I thought it was ok, but got a message from another member saying it might not have been) and you've now confirmed it isn't. Thanks for letting me know
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    (Original post by MathMeister)
    I actually done most of the thinking for myself. I just needed help really with the roots thing. I don't see how anybody can figure out that real roots means equal and different. Everybody learns that. It's not something you can figure out. I also done the full working, up to and including the inequalities and graph. Just backin' y'all up.
    Ok thanks - I saw that, in your question, you went along with your method of finding the quadratic etc.

    My working out was basically just showing you that you had indeed gone along the correct pathway, just that we'd got it a little bit wrong with the inequalities

    Glad I could help you If you have any other questions on which you are totally stuck on, I might drop in a few hints, but I'm not risking getting a nice collection of cards :nah:
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    (Original post by spotify95)
    Okay thanks - I was wondering whether posting full solutions was allowed on here or not (I thought it was ok, but got a message from another member saying it might not have been) and you've now confirmed it isn't. Thanks for letting me know
    Yeah, like TenOfThem said, we don't allow any full solutions, thanks for understanding.
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    (Original post by MathsNerd1)
    Yeah, like TenOfThem said, we don't allow any full solutions, thanks for understanding.
    No problem, thanks for letting me know. I'd rather you tell me that they aren't allowed, than get a card of some sort for it :yep:

    I'll stick to just posting hints then, to get the user to work out the right answer.
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    (Original post by MathMeister)
    Alas, I have another question in dire need of answering. Please may somebody help. 4a) Find the set of values for k for which the equation x^2-kx+(k+3)=0 has real roots [...] First of all I don't understand what the question is asking for [...] It's asking for what part of the equation has real roots, but the whole equation- the equation in itself has real roots, which is why I am confused with what they are asking me for. Please may somebody help?! Ahh wait, it's asking for the range of values for k in which the equation would satisfy the b squared = 4ac. But I do not understand. Brb. Going out. Please help explain the question and answer.

    I'll have a go at explaining the question.

    Think of it like this:

    x^2-kx+(k+3)=0 can become different equations depending on what k equals.

    e.g. if k=0, it becomes x^2+3=0
    This has no real roots (i.e. no real solutions – you can think of the word "root" as meaning "solution" in this context), which can easily be seen by rearranging it to x^2=-3

    Imagine setting k equal to different numbers, seeing what the equation becomes, and deciding if it has any real roots.

    If k=1 the equation becomes x^2-x+4=0 (no real roots)
    If k=2 the equation becomes x^2-2x+5=0 (no real roots)
    If k=6 the equation becomes x^2-6x+9=0 (aha, a real root i.e. 3)
    If k=7 the equation becomes x^2-7x+10=0 (real roots i.e. 2 or 5)

    So some values of k turn x^2-kx+(k+3)=0 into an equation which has real roots and other values of k turn it into an equation which doesn't have real roots.

    The question is asking you to find the complete set of values of k which turn it into an equation with real roots (which, as posters have pointed out, is under -2 or over 6 inclusive).
 
 
 
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