Hey there! Sign in to join this conversationNew here? Join for free
x Turn on thread page Beta
    • Thread Starter
    Offline

    19
    ReputationRep:
    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?
    • Community Assistant
    • Study Helper
    Offline

    20
    ReputationRep:
    Community Assistant
    Study Helper
    (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

    a(x-p)(x-q)

    where p and q are the roots of the quadratic.

    Expand this and compare it with ax^2+bx+c.
    • Community Assistant
    • Study Helper
    Offline

    20
    ReputationRep:
    Community Assistant
    Study Helper
    (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 \frac{c}{a}

    and the sum is equal to -\frac{b}{a}.
    Offline

    20
    ReputationRep:
    you could also try adding and multiplying the two versions ( one with +, one with - ) of the quadratic formula
    • Thread Starter
    Offline

    19
    ReputationRep:
    (Original post by notnek)
    Also a correction:

    The product of the two roots is equal to \frac{c}{a}

    and the sum is equal to -\frac{b}{a}.
    Yeah sorry, I knew that, I was meant to write when a is <1 .
    • Thread Starter
    Offline

    19
    ReputationRep:
    (Original post by notnek)
    A quadratic can be written in the form

    a(x-p)(x-q)

    where p and q are the roots of the quadratic.

    Expand this and compare it with ax^2+bx+c.
    Is that the actual way to prove the theorem - that the products are equal to +c
    • Community Assistant
    • Study Helper
    Offline

    20
    ReputationRep:
    Community Assistant
    Study Helper
    (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.
    • Thread Starter
    Offline

    19
    ReputationRep:
    (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!
    • Thread Starter
    Offline

    19
    ReputationRep:
    (Original post by the bear)
    you could also try adding and multiplying the two versions ( one with +, one with - ) of the quadratic formula
    Thanks
    • Thread Starter
    Offline

    19
    ReputationRep:
    (Original post by notnek)
    A quadratic can be written in the form

    a(x-p)(x-q)

    where p and q are the roots of the quadratic.

    Expand this and compare it with ax^2+bx+c.
    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
    Offline

    12
    ReputationRep:
    (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.
    • Thread Starter
    Offline

    19
    ReputationRep:
    (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.


    Posted from TSR Mobile
    Offline

    12
    ReputationRep:
    (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  ax^2 + bx + c \equiv ax^2 - apx - aqx + apq {Note that the use of  \equiv 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  - ap - aq = b and hence  p + q = \frac{-b}{a} . We would not write  -apx - aqx = bx , we would only write the coefficients down.
    • Thread Starter
    Offline

    19
    ReputationRep:
    (Original post by 16Characters....)
    You are correct that  ax^2 + bx + c \equiv ax^2 - apx - aqx + apq {Note that the use of  \equiv 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  - ap - aq = b and hence  p + q = \frac{-b}{a} . We would not write  -apx - aqx = bx , we would only write the coefficients down.
    Oh right lol thanks


    Posted from TSR Mobile
    • Thread Starter
    Offline

    19
    ReputationRep:
    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:

    (x-1)^3 = (x-1)(1+x)

(x^2-2x+1)(x-1) = (x-1)(1+x)

(x^3 - 2x^2 + x) - (x^2 - 2x + 1) = (x-1)(1+x)

x^3 - 3x^2 + 3x - 1 = x^2 + x - x - 1

x^3 - 3x^2 + 3x - 1 = x^2 - 1

x^3 - 4x^2 + 3x = 0

x(x^2 - 4x + 3) = 0

x(x-3)(x-1) = 0

So, x = 0 or x = 3 or x = 1.

    I think this part of the method is pretty straightforward and fast.

    y = (x-1)(1+x)



When x = 0, y = (0-1)(1+0)

= (-1)(1)

= -1



When x = 3, y = (3-1)(1+3)

= (2)(4)

= 8



When x = 1, y = (1-1)(1+1)

= (0)(2)

= 0



So, the co-ordinates of the points of intersection are (0,-1), (3,8) and (1,0).
    • Very Important Poster
    Offline

    21
    ReputationRep:
    Very Important Poster
    (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.
    • Thread Starter
    Offline

    19
    ReputationRep:
    (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?
    • Very Important Poster
    Offline

    21
    ReputationRep:
    Very Important Poster
    (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).
    • Thread Starter
    Offline

    19
    ReputationRep:
    (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).
    So, (x-1)^3 - (x-1)(1+x) = 0

(x-1)^2(1+x) = 0

(x^2-2x+1)(1+x) = 0

(x^3 - 2x^2 + x) + (x^2 - 2x + 1) = 0

x^3 - x^2 - x + 1 = 0

    This is clearly incorrect.

    V2

    (x-1)^3 - (x-1)(1+x) = 0

(x-1)^2 - (1+x) = 0

(x^2 - 2x + 1) - 1 - x = 0

x^2 - 3x = 0

x(x-3) = 0

x = 0 or x = 3

    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.
    • Very Important Poster
    Offline

    21
    ReputationRep:
    Very Important Poster
    (Original post by Chittesh14)
    So, (x-1)^3 - (x-1)(1+x) = 0

(x-1)^2(1+x) = 0

(x^2-2x+1)(1+x) = 0

(x^3 - 2x^2 + x) + (x^2 - 2x + 1) = 0

x^3 - x^2 - x + 1 = 0

    This is clearly incorrect.

    V2

    (x-1)^3 - (x-1)(1+x) = 0

(x-1)^2 - (1+x) = 0

(x^2 - 2x + 1) - 1 - x = 0

x^2 - 3x = 0

x(x-3) = 0

x = 0 or x = 3

    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.
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: November 21, 2015
Poll
Do you agree with the proposed ban on plastic straws and cotton buds?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.