Hello guys,
I was wondering that since the product of the two roots of a quadratic equation equal to the yintercept (c) in the quadratic form and that the sum of the two roots equal to (b) in the quadratic form.
Does anyone know why or how this was invented?

Chittesh14
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 24102015 18:07

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 24102015 18:15
(Original post by Chittesh14)
Hello guys,
I was wondering that since the product of the two roots of a quadratic equation equal to the yintercept (c) in the quadratic form and that the sum of the two roots equal to (b) in the quadratic form.
Does anyone know why or how this was invented?
where and are the roots of the quadratic.
Expand this and compare it with . 
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 24102015 18:19
(Original post by Chittesh14)
Hello guys,
I was wondering that since the product of the two roots of a quadratic equation equal to the yintercept (c) in the quadratic form and that the sum of the two roots equal to (b) in the quadratic form.
Does anyone know why or how this was invented?
The product of the two roots is equal to
and the sum is equal to . 
the bear
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 24102015 18:21
you could also try adding and multiplying the two versions ( one with +, one with  ) of the quadratic formula

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 24102015 18:31
(Original post by notnek)
Also a correction:
The product of the two roots is equal to
and the sum is equal to . 
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 24102015 18:32
(Original post by notnek)
A quadratic can be written in the form
where and are the roots of the quadratic.
Expand this and compare it with . 
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 24102015 18:35
(Original post by Chittesh14)
Is that the actual way to prove the theorem  that the products are equal to +c 
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 24102015 18:36
(Original post by notnek)
It's one way. The bear has given you another way. 
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 24102015 18:37
(Original post by the bear)
you could also try adding and multiplying the two versions ( one with +, one with  ) of the quadratic formula 
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 24102015 19:48
(Original post by notnek)
A quadratic can be written in the form
where and are the roots of the quadratic.
Expand this and compare it with .
a(x^2xqpx+pq)
ax^2  axq  apx + apq = 0
ax^2 + bx + c = 0
apq = c
pq = c/a
 axq  apx = bx
 xq  px = bx/a
 q  p = b/a
q + p = b/a
Is this correct?
Done in head and on iPad so sorry if there are any spelling mistakes lol.
Posted from TSR MobileLast edited by Chittesh14; 24102015 at 19:50. 
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 24102015 20:09
(Original post by Chittesh14)
a(xp)(xq)
a(x^2xqpx+pq)
ax^2  axq  apx + apq = 0
ax^2 + bx + c = 0
apq = c
pq = c/a
 axq  apx = bx
 xq  px = bx/a
 q  p = b/a
q + p = b/a
Is this correct?
Done in head and on iPad so sorry if there are any spelling mistakes lol.
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As an exercise you might now try to find similar results for a cubic equation.Last edited by 16Characters....; 24102015 at 20:14. 
Chittesh14
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 24102015 21:14
(Original post by 16Characters....)
Edit: Actually not quite. You can compare the coefficients, e.g. p  q = b/a etc but not the complete terms (i.e. no px  qx = bx/a).
As an exercise you might now try to find similar results for a cubic equation.
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 24102015 21:20
(Original post by Chittesh14)
Sorry I don't get what you mean by the first line :/, especially the complete terms part.
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From here we can do what is called "comparing coefficients" which is exactly what is says on the tin. We can compare the coefficients of x to see that and hence . We would not write , we would only write the coefficients down. 
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 24102015 22:02
(Original post by 16Characters....)
You are correct that {Note that the use of means that it holds whatever value of x you substitute in. It means "is the same as" or "is equivalent to"}
From here we can do what is called "comparing coefficients" which is exactly what is says on the tin. We can compare the coefficients of x to see that and hence . We would not write , we would only write the coefficients down.
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 21112015 15:17
Can anyone please tell me if there is a shorter way of working out this question ?
There are two curves y = (x1)^3 and y = (x1)(1+x).
I need to find the coordinates of the points of intersection.
This is my method:
I think this part of the method is pretty straightforward and fast.
Last edited by Chittesh14; 21112015 at 15:20. 
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 21112015 15:24
(Original post by Chittesh14)
x
Just to make it easier for you to see, if we swapped x1 for a then we would have a^3 = a(1+x).
So then a^3  a(1+x) = 0 and there is a common factor between those two terms.
That saves you a bit of expanding. 
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 21112015 16:29
(Original post by SeanFM)
Yes.
Just to make it easier for you to see, if we swapped x1 for a then we would have a^3 = a(1+x).
So then a^3  a(1+x) = 0 and there is a common factor between those two terms.
That saves you a bit of expanding.
x^3  x(1+x) = 0
x^3  x^2  x = 0
x(x^2  x  1) = 0
Then what? 
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 21112015 16:42
(Original post by Chittesh14)
So, x^3 = x(1+x).
x^3  x(1+x) = 0
x^3  x^2  x = 0
x(x^2  x  1) = 0
Then what?
a^3  a(1+x) = 0
You've then multiplied it out and then factored it out which is fine but you can skip the third line by just noticing that a is the commmon factor, so a^3  a(1+x) = a(a^2  (1+x)), and then substitute a back out and factorise whatever's in the brackets. Ideally you don't need to use 'a', just use (x1) from the start and notice that you can factor out (x1) from (x1)^3  (x1)(x+1). 
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 21112015 16:57
(Original post by SeanFM)
Oh, you should keep x1 in there instead of putting a (which you've now changed into x which is very dangerous!). I just labelled it a to make it clear that you were factoring out.
a^3  a(1+x) = 0
You've then multiplied it out and then factored it out which is fine but you can skip the third line by just noticing that a is the commmon factor, so a^3  a(1+x) = a(a^2  (1+x)), and then substitute a back out and factorise whatever's in the brackets. Ideally you don't need to use 'a', just use (x1) from the start and notice that you can factor out (x1) from (x1)^3  (x1)(x+1).
This is clearly incorrect.
V2
Partially correct, where is the x = 1 gone?
I know that if I multiply the equation above by (x1), I'll get the correct answer, but I don't know how to bring that (x1) back.Last edited by Chittesh14; 21112015 at 17:01. 
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 21112015 17:01
(Original post by Chittesh14)
This is clearly incorrect.
V2
Partially correct, where is the x = 1 gone?
For V2 you're almost there, but you 'cancelled out' x1 from going to the first line to the second line. (if you put each line in brackets from line 2 onwards and multiply by (x1) then you will get what you did by going through the long way.
You shouldn't cancel out solutions, instead just factorise them. Like if you have sinx = tanx, then you would say sinx = (sinx)/cosx and it would be a mistake to divide both sides by sinx as you would be getting rid of a solution.
Instead, you go cosxsinx  sinx = 0 and so sinx(cosx  1) = 0. And your question is similar to this.
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