The title says it all. This is an inequality. I came up with x is smaller than 2 and bigger than 2.5 and 3 however the answer is x>3. But this excludes the 2 and the 2.5 Please help.
and

MathMeister
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 13072014 13:50

MathMeister
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 13072014 13:55
Ahhh I see, I just needed to see them as graphs. x<2.5 I should've seen as a graph. Graphs are so helpful. I got the right answer now.
Btw, does anybody know whether you can, and if so, how to delete threads you've posted? 
TenOfThem
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 13072014 13:55
(Original post by MathMeister)
The title says it all. This is an inequality. I came up with x is smaller than 2 and bigger than 2.5 and 3 however the answer is x>3. But this excludes the 2 and the 2.5 Please help.
and
2x>5 means that you cannot have any values less than 2.5
Then from the first equation you have
x cannot be between 2 and 3
So this now tells you x>3 
TenOfThem
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 13072014 13:56
(Original post by MathMeister)
how to delete threads you've posted? 
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 13072014 14:03
(Original post by MathMeister)
...
These are both satisfied bySpoiler:Show(Original post by TenOfThem)
...Last edited by Khallil; 13072014 at 14:22. 
MathMeister
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 13072014 14:11
I see the importance like many have said of learning/revising hard from the modular books instead of smashing exam papers as a excuse for learning. However, I have recently became even more confused, exactly because they have to satisfy each inequality. And just as I write this I figure it out. Gosh the textbook is so vital in learning.

spotify95
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 13072014 14:30
(Original post by MathMeister)
The title says it all. This is an inequality. I came up with x is smaller than 2 and bigger than 2.5 and 3 however the answer is x>3. But this excludes the 2 and the 2.5 Please help.
and
For the quadratic, just factorise as normal  gives crossing points as or . However, since the inequality says "greater than 0", you want all solutions that are greater than 0. It's easiest to graph a sketch. You'll realise that the bit that is below zero is the region: . Hence the rest of it is the first solution, i.e. and .
This then gets combined with the other inequality, which works out to be . However, when combined, we know that x must be larger than 3. Hence, x can no longer be less than 2, or between 2.5 and 3. This leads to the correct solution, . 
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 13072014 14:33
(Original post by TenOfThem)
there is no need to delete  others may learn from your initial confusion
Subscribers can delete their own threads, or alternatively, go to Ask A Moderator and one of the mods can delete it for you. 
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 13072014 14:57
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 has real roots. I thankfully knew real roots meant that it had the discriminant b squared was greater or equal to 4 ac. So I subbed the parts of the equation into the b squared = 4ac thing and solved for k. I then hoped/ and got a quadratic which meant I could get the answer, however I have a few quetions about the answer; 2<k<6. First of all I don't understand what the question is asking for. I understand the dicriminant, roots , recurring roots knowledge is important for a variety of questions, however I am not sure what the question was 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.

spotify95
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 13072014 15:04
(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 has real roots. I thankfully knew real roots meant that it had the discriminant b squared was greater or equal to 4 ac. So I subbed the parts of the equation into the b squared = 4ac thing and solved for k. I then hoped/ and got a quadratic which meant I could get the answer, however I have a few questions about the answer; 2<k<6. First of all I don't understand what the question is asking for. I understand the dicriminant, roots , recurring roots knowledge is important for a variety of questions, however I am not sure what the question was 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.
Hi,
I'll give it a crack now. It is asking for the set of values for k, that let 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: .
By plugging in the values for a, b and c, and simplifying, we get:
Solving gives the roots of the quadratic as or .
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:
and
I've also attached my methodology to this problem.
Hope I've helpedLast edited by spotify95; 13072014 at 15:25. 
TenOfThem
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 13072014 15:43
(Original post by spotify95)
Hi 
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 13072014 16:45
(Original post by TenOfThem)
I am very surprised to see a full solution from someone who officially represents TSR
The reason why I posted it was to explain how I got to that answer. It was more of a confirmation, as the FM that asked me the question did get somewhere along the lines of the correct answer (mentioned a quadratic in k). The only thing he got wrong was the inequality sign; he answered 2 < k < 6 when the answer is k < 2 and k > 6.
Graphing this type of question always helps  but don't worry I'll edit if necessary.Last edited by spotify95; 13072014 at 16:50. 
MathMeister
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 13072014 18:20
What I don't understand is why it isn't greater or equal instead of just being greaterSince when b squared = 4ac, there are roots, two of them, which happen to be equal and ''recurring'' as well as b squared being larger than 4ac having real roots. So why isn't b^2 greater or equal to 4ac in the calculation, or is it something to do with the graph?
Also, in the answer book in the modular book and cd, the answer is 2<k<6. ???? Misprint/mistake? This is rather interesting. GeoGebra also agrees with you. I think its a misprint. possibly
Also, what's with the cross in your zero?Last edited by MathMeister; 13072014 at 18:33. 
spotify95
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 13072014 18:59
(Original post by MathMeister)
What I don't understand is why it isn't greater or equal instead of just being greaterSince when b squared = 4ac, there are roots, two of them, which happen to be equal and ''recurring'' as well as b squared being larger than 4ac having real roots. So why isn't b^2 greater or equal to 4ac in the calculation, or is it something to do with the graph?
Also, in the answer book in the modular book and cd, the answer is 2<k<6. ???? Misprint/mistake? This is rather interesting. GeoGebra also agrees with you. I think its a misprint. possibly
Also, what's with the cross in your zero?
I assumed you wanted real roots, and it is greater than when there are real roots.
You are correct in that it can be equal to zero; however then they are called "equal roots", as the root is the same (it is a touching point rather than a crossing point on the curve)
It probably is a misprint  as 2<k<6 would be right for if it was less than zero. However, less than zero there are no real roots, so that is not possible. I would say a mis print.
The cross in my zero? You mean the line? Probably how I write them, I can make it clearer for you from now on 
MathMeister
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 13072014 19:41
So, to conclude this question, a real root is when there are 2 xintercepts. A real root cannot be an equal root (vice versa) as an equal root does not cross the xaxis. 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! 
Arithmeticae
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 13072014 19:44
(Original post by MathMeister)
So, to conclude this question, a real root is when there are 2 xintercepts. A real root cannot be an equal root (vice versa) as an equal root does not cross the xaxis. 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!
A real root is just that, it doesn't matter if it's repeated or two distinct roots.
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TenOfThem
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 13072014 19:46
(Original post by MathMeister)
So, to conclude this question, a real root is when there are 2 xintercepts. A real root cannot be an equal root (vice versa) as an equal root does not cross the xaxis. Am I correct in thinking this?
a real root is Real rather than Complex
A quadratic can have no Real roots
 a repeated real root
 2 real roots

Arithmeticae
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 13072014 19:49
(Original post by MathMeister)
There are many contradictions. Maybe TenOfThem may hopefully remove all confusion. So many different opinions...
If you are right then spotify95 was wrong in not putting greater than or equal.
Real roots  ∆ => 0
Distinct real roots  ∆ >0
Repeated real roots  ∆=0
Complex roots  ∆ <0Last edited by Arithmeticae; 13072014 at 19:56. 
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 13072014 19:54
Prsom* ... to allIn 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 arithmatic posted (ty) 
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 13072014 20:02
(Original post by MathMeister)
Prsom* ... to allIn 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 arithmatic posted (ty)
The words Real Roots implies more than 1 real root
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