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Can you explain this to me (involves real roots and completing the square)? watch

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    (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 = 4-c

    But it then goes to 4-c > 0.

    What happened to the (x+2)^2? Where did that go? And why is it >0?

    Thanks
    So you understand how they got (x+2)^2=4-c

    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 (x+2)^2>0 and since we know know (x+2)^2 and 4-c are equal to eachother, ie the same thing, then we can replace the LHS of the inequality with 4-c hence 4-c>0
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    (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 = 4-c] and then gets to the step of 4-c>0.

    Can you explain how they did that please?
    Its applying what i said above but i think its just skipping steps as -(4-c)<0 is 4-c>0

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    (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 = 4-c] and then gets to the step of 4-c>0.

    Can you explain how they did that please?
    I think the easiest way to understand this is to think about the possible values of  (x+2)^2 . Something squared is always equal to or greater than 0, so  (x+2)^2 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  (x+2)^2&gt;0 This has been worked out just be considering that side of the equation
    You know 4-c is equal to this, so 4-c must be greater than 0 as well.
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    (Original post by RDKGames)
    So you understand how they got (x+2)^2=4-c

    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 (x+2)^2&gt;0 and since we know know (x+2)^2 and 4-c are equal to eachother, ie the same thing, then we can replace the LHS of the inequality with 4-c hence 4-c&gt;0
    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 = 4-c?

    And secondly, when you get to the stage of (x+2)^2 = 4-c where, as you said, it needs to be b^2-4ac>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 4-c on the right hand side), how do you know that the 4-c is the discriminant?

    I mean, with the line (x+2)^2 = 4-c, I get that the LHS must be >0, so you can replace the LHS with 4-c because both sides of the equation are equal to each other. However, I don't get how you know to relate that 4-c on the LHS to being the discriminant (so you know the 4-c must be >0). How do you know that the 4-c on the LHS is the discriminant?

    Is it just that when you replaced the equation (x+2)^2 = 4-c with just 4-c=0, you knew that the LHS, the 4-c must be greater than 0 for the condition of unequal real roots to be true, so you changed the 4-c=0 into 4-c>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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    (Original post by sindyscape62)
    I think the easiest way to understand this is to think about the possible values of  (x+2)^2 . Something squared is always equal to or greater than 0, so  (x+2)^2 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  (x+2)^2&gt;0 This has been worked out just be considering that side of the equation
    You know 4-c is equal to this, so 4-c must be greater than 0 as well.
    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 4-c 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 4-c>0, how do you know that that is the same as the discriminant, b^2-4ac, being >0? I would never have been able to spot 4-c as the discriminant and know that therefore, 4-c must be >0?

    Is it just that once you get the 4-c>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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    (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 = 4-c?
    Because in that form you can consider (x+2)^2 and c-4 seperately - which is important because one side is a square of a number and this gives us information about what values it can take up, hence we get information on what 4-c can be. So for example, the LHS cannot be negative. And if it is 0 then we have equal roots, etc... as I mentioned previously.

    And secondly, when you get to the stage of (x+2)^2 = 4-c where, as you said, it needs to be b^2-4ac>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 4-c on the right hand side), how do you know that the 4-c is the discriminant?
    I did not say that, the discriminant is a different approach and I did not mention it. The discriminant is b^2-4ac for a quadratic in the form ax^2+bx+c=0 and since we do not have that form here, we cannot use the discriminant directly. Of course, if you wish, you may get it into the expanded form and do it from there, I don't think it matters unless the question specifically says to derive it from the completed square form.

    I mean, with the line (x+2)^2 = 4-c, I get that the LHS must be >0, so you can replace the LHS with 4-c because both sides of the equation are equal to each other. However, I don't get how you know to relate that 4-c on the LHS to being the discriminant (so you know the 4-c must be >0). How do you know that the 4-c on the LHS is the discriminant?
    4-c>0 is the simplified discriminant because the actual discriminant is 16-4c.

    Is it just that when you replaced the equation (x+2)^2 = 4-c with just 4-c=0, you knew that the LHS, the 4-c must be greater than 0 for the condition of unequal real roots to be true, so you changed the 4-c=0 into 4-c>0? Am I correct in thinking this? Is that how you knew that the LHS of the equation was the discriminant?
    Again, this is not about the discriminant as you would find it from a general quadratic form. I considered LHS first before I considered RHS so I was not applying the conditions on 4-c straight away. You would find that this is the exact value of the discriminant if you were to plug in values for it from the expanded form, but we are not taking this common approach to the problem and are instead thinking about it from a different perspective.
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    (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 4-c 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 4-c>0, how do you know that that is the same as the discriminant, b^2-4ac, being >0? I would never have been able to spot 4-c as the discriminant and know that therefore, 4-c must be >0?

    Is it just that once you get the 4-c>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?
    Don't confuse 4-c with the discriminant- you're talking about two different methods which will both get you the right answer
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    (Original post by RDKGames)
    So you understand how they got (x+2)^2=4-c

    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 (x+2)^2&gt;0 and since we know know (x+2)^2 and 4-c are equal to eachother, ie the same thing, then we can replace the LHS of the inequality with 4-c hence 4-c&gt;0
    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?
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    (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.
    Roots are those which satisfy the quadratic equaling 0. Anything to do with the quantity and nature of roots is down to the determinant being some value on either side of 0.

    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?
    I know they have to be real because this is a C1 type question, and it says so in the question so I absolutely cannot have complex numbers. Strictly speaking, the question must specify that the roots are real otherwise you get unequal roots when 4-c\not= 0 if you bring in the complex numbers.

    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?
    Refer to my answer above. Basically the question must outline the conditions.
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    ANY square real number is greater than or equal to 0, right? So since  (x+2)^2 is a square number we must have, by the same argument that  \underline{(x+2)^2\geq 0} , right?

    Now since  (x+2)^2=4-c we can replace the  (x+2)^2 in the underlined inequality with  4-c right since they are the same.
    So we have  4-c\geq 0 . Now if  4-c=0 we have equal roots, because if  4-c=(x+2)^2=0 then  x+2=0 which only gives one (2 repeated roots) root which would be  x=-2 right?
    So we conclude that  4-c\neq 0 and so we have  4-c&gt;0 .
 
 
 
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