So you understand how they got(Original post by blobbybill)
The question is:
"Given that the equation x^2 + 4x + c =0 has unequal real roots, find the range of possible values of c"
From earlier on in the question, I know that x^2 + 4x + c = (x+2)^2 4 + c =0
I get the (x+2)^2 4 + c =0 because the questions earlier stages involved completing the square.
However, in the solution, (x+2)^2 4 + c =0 then goes to (x+2)^2 = 4c
But it then goes to 4c > 0.
What happened to the (x+2)^2? Where did that go? And why is it >0?
Thanks
Now for unequal roots, you know that the LHS cannot be less than 0 otherwise you have imaginary roots. We know it cannot be =0 because otherwise we get equal roots. So the only option left is if it is greater that 0 hence our condition goes to and since we know know and are equal to eachother, ie the same thing, then we can replace the LHS of the inequality with 4c hence

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 06112016 19:29

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 06112016 19:32
(Original post by blobbybill)
I get and understand that. What I don't get, is how the markscheme goes from the step on the second line [(x+2)^2 = 4c] and then gets to the step of 4c>0.
Can you explain how they did that please?
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 06112016 19:37
(Original post by blobbybill)
I get and understand that. What I don't get, is how the markscheme goes from the step on the second line [(x+2)^2 = 4c] and then gets to the step of 4c>0.
Can you explain how they did that please?
You know 4c is equal to this, so 4c must be greater than 0 as well. 
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 06112016 19:37
(Original post by RDKGames)
So you understand how they got
Now for unequal roots, you know that the LHS cannot be less than 0 otherwise you have imaginary roots. We know it cannot be =0 because otherwise we get equal roots. So the only option left is if it is greater that 0 hence our condition goes to and since we know know and are equal to eachother, ie the same thing, then we can replace the LHS of the inequality with 4c hence
However, when they got (x+2)^2 4 +c =0, what is the reasoning behind moving the 4 and +c to the RHS in order to get to (x+2)^2 = 4c?
And secondly, when you get to the stage of (x+2)^2 = 4c where, as you said, it needs to be b^24ac>0 and the LHS must be >0 in order to fulfil the condition, how do you know that the LHS of the equation (or the 4c on the right hand side), how do you know that the 4c is the discriminant?
I mean, with the line (x+2)^2 = 4c, I get that the LHS must be >0, so you can replace the LHS with 4c because both sides of the equation are equal to each other. However, I don't get how you know to relate that 4c on the LHS to being the discriminant (so you know the 4c must be >0). How do you know that the 4c on the LHS is the discriminant?
Is it just that when you replaced the equation (x+2)^2 = 4c with just 4c=0, you knew that the LHS, the 4c must be greater than 0 for the condition of unequal real roots to be true, so you changed the 4c=0 into 4c>0? Am I correct in thinking this? Is that how you knew that the LHS of the equation was the discriminant? 
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 06112016 19:44
(Original post by sindyscape62)
I think the easiest way to understand this is to think about the possible values of . Something squared is always equal to or greater than 0, so is equal to or greater than 0. You know it can't be 0 because the question says there are two different roots x=2 is the only value which would make it 0, so there'd be a repeated root. This means This has been worked out just be considering that side of the equation
You know 4c is equal to this, so 4c must be greater than 0 as well.
I understand all of what you just said, and am I also right in saying you know the 4c and the (x+2)^2 must be >0 because you are told the equation has unreal equal roots, so it must be >0? Is that correct?
Anyway, once you get to the point where you have 4c>0, how do you know that that is the same as the discriminant, b^24ac, being >0? I would never have been able to spot 4c as the discriminant and know that therefore, 4c must be >0?
Is it just that once you get the 4c>0, you have something that is <0, =0, or >0, so you can then make that link in your head and recognise that it is the discriminant that is>0? 
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 06112016 19:45
(Original post by blobbybill)
That clears it up a lot, thank you.
However, when they got (x+2)^2 4 +c =0, what is the reasoning behind moving the 4 and +c to the RHS in order to get to (x+2)^2 = 4c?
And secondly, when you get to the stage of (x+2)^2 = 4c where, as you said, it needs to be b^24ac>0 and the LHS must be >0 in order to fulfil the condition, how do you know that the LHS of the equation (or the 4c on the right hand side), how do you know that the 4c is the discriminant?
I mean, with the line (x+2)^2 = 4c, I get that the LHS must be >0, so you can replace the LHS with 4c because both sides of the equation are equal to each other. However, I don't get how you know to relate that 4c on the LHS to being the discriminant (so you know the 4c must be >0). How do you know that the 4c on the LHS is the discriminant?
Is it just that when you replaced the equation (x+2)^2 = 4c with just 4c=0, you knew that the LHS, the 4c must be greater than 0 for the condition of unequal real roots to be true, so you changed the 4c=0 into 4c>0? Am I correct in thinking this? Is that how you knew that the LHS of the equation was the discriminant?Last edited by RDKGames; 06112016 at 19:51. 
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 06112016 19:51
(Original post by blobbybill)
That clears it up too, thank you.
I understand all of what you just said, and am I also right in saying you know the 4c and the (x+2)^2 must be >0 because you are told the equation has unreal equal roots, so it must be >0? Is that correct?
Anyway, once you get to the point where you have 4c>0, how do you know that that is the same as the discriminant, b^24ac, being >0? I would never have been able to spot 4c as the discriminant and know that therefore, 4c must be >0?
Is it just that once you get the 4c>0, you have something that is <0, =0, or >0, so you can then make that link in your head and recognise that it is the discriminant that is>0? 
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 06112016 19:54
(Original post by RDKGames)
So you understand how they got
Now for unequal roots, you know that the LHS cannot be less than 0 otherwise you have imaginary roots. We know it cannot be =0 because otherwise we get equal roots. So the only option left is if it is greater that 0 hence our condition goes to and since we know know and are equal to eachother, ie the same thing, then we can replace the LHS of the inequality with 4c hence
On another note, this is just to clear the last drops of confusion I have, when you say that "you know that the LHS cannot be less than 0 otherwise you have imaginary roots. We know it cannot be =0 because otherwise we get equal roots.", how do you know that we cannot have equal roots in this question, and that the answer wants it to have real roots?
This is a really daft question, but do you know that you need to make it have real roots and be >0 because the question says that the equation has real roots? So if it says it has real roots in the question, you know it must be >0? Was that the only way you knew the LHS was >0?
If it wasn't, how else, other than the question saying it had real roots, did you know that it had to be >0? 
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 06112016 20:01
(Original post by blobbybill)
Ah, ok. I had (wrongly) thought that the LHS being =0, <0, >0 only applied when using the discriminant. I had never realised that it just related to the LHS of an equation being either =0, <0 or >0. So basically, we know that the LHS of the equation is >0, because otherwise we would get equal roots or imaginary roots. So it is a "root" because we have 0 on the RHS, and we have the LHS > 0, so we know it has unequal real roots.
On another note, this is just to clear the last drops of confusion I have, when you say that "you know that the LHS cannot be less than 0 otherwise you have imaginary roots. We know it cannot be =0 because otherwise we get equal roots.", how do you know that we cannot have equal roots in this question, and that the answer wants it to have real roots?
This is a really daft question, but do you know that you need to make it have real roots and be >0 because the question says that the equation has real roots? So if it says it has real roots in the question, you know it must be >0? Was that the only way you knew the LHS was >0?
If it wasn't, how else, other than the question saying it had real roots, did you know that it had to be >0?Last edited by RDKGames; 06112016 at 20:05. 
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 06112016 20:18
ANY square real number is greater than or equal to 0, right? So since is a square number we must have, by the same argument that , right?
Now since we can replace the in the underlined inequality with right since they are the same.
So we have . Now if we have equal roots, because if then which only gives one (2 repeated roots) root which would be right?
So we conclude that and so we have .
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