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# Question regarding quadratic equation roots watch

1. Hello guys,

I was wondering that since the product of the two roots of a quadratic equation equal to the y-intercept (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?
2. (Original post by Chittesh14)
Hello guys,

I was wondering that since the product of the two roots of a quadratic equation equal to the y-intercept (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?
A quadratic can be written in the form

where and are the roots of the quadratic.

Expand this and compare it with .
3. (Original post by Chittesh14)
Hello guys,

I was wondering that since the product of the two roots of a quadratic equation equal to the y-intercept (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?
Also a correction:

The product of the two roots is equal to

and the sum is equal to .
4. you could also try adding and multiplying the two versions ( one with +, one with - ) of the quadratic formula
5. (Original post by notnek)
Also a correction:

The product of the two roots is equal to

and the sum is equal to .
Yeah sorry, I knew that, I was meant to write when a is <1 .
6. (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 .
Is that the actual way to prove the theorem - that the products are equal to +c
7. (Original post by Chittesh14)
Is that the actual way to prove the theorem - that the products are equal to +c
It's one way. The bear has given you another way.
8. (Original post by notnek)
It's one way. The bear has given you another way.
Thanks :P. I never knew it was that easy lol!
9. (Original post by the bear)
you could also try adding and multiplying the two versions ( one with +, one with - ) of the quadratic formula
Thanks
10. (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-p)(x-q)
a(x^2-xq-px+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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11. (Original post by Chittesh14)
a(x-p)(x-q)
a(x^2-xq-px+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 Mobile
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.
12. (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.
Sorry I don't get what you mean by the first line :/, especially the complete terms part.

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13. (Original post by Chittesh14)
Sorry I don't get what you mean by the first line :/, especially the complete terms part.

Posted from TSR Mobile
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.
14. (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.
Oh right lol thanks

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15. Can anyone please tell me if there is a shorter way of working out this question ?

There are two curves y = (x-1)^3 and y = (x-1)(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.

16. (Original post by Chittesh14)
x
Yes.

Just to make it easier for you to see, if we swapped x-1 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.
17. (Original post by SeanFM)
Yes.

Just to make it easier for you to see, if we swapped x-1 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.
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?
18. (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?
Oh, you should keep x-1 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 (x-1) from the start and notice that you can factor out (x-1) from (x-1)^3 - (x-1)(x+1).
19. (Original post by SeanFM)
Oh, you should keep x-1 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 (x-1) from the start and notice that you can factor out (x-1) from (x-1)^3 - (x-1)(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 (x-1), I'll get the correct answer, but I don't know how to bring that (x-1) back.
20. (Original post by Chittesh14)

This is clearly incorrect.

V2

Partially correct, where is the x = 1 gone?
For V1 I'm not quite sure how you got to your second line.

For V2 you're almost there, but you 'cancelled out' x-1 from going to the first line to the second line. (if you put each line in brackets from line 2 onwards and multiply by (x-1) 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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