Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    0
    ReputationRep:
    I just need to know what to do :/

    The cubic equation x^3 + Ax^2 + Bx + 15 = 0

    Where A and B = real numbers

    Root = x = 1+2j

    Q. Find the value of the real root and the values of A and B

    I did something weird and got A as 2j and B as 4, but I heard the correct answer is 1 and -1, can someone explain the process? Thanks
    • Community Assistant
    Offline

    20
    ReputationRep:
    Community Assistant
    If the coefficient of a polynomial are real then any complex roots must occur in conjugate pairs. This means you can write down a second root. You now have two values you can put in the equation for x and this will enable you to obtain a pair of simultaneous equations in A and B.
    • Thread Starter
    Offline

    0
    ReputationRep:
    I know the second root is x = 1-2j, but I have no idea what to do from there :/
    Offline

    1
    ReputationRep:
    if one root is 1+2j then another is 1-2j due to complex conjugates. Therefore factors will be x-1-2j and x-1+2j . Multiply them together and do a long division to find the third factor.
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by superfoggy)
    if one root is 1+2j then another is 1-2j due to complex conjugates. Therefore factors will be x-1-2j and x-1+2j . Multiply them together and do a long division to find the third factor.
    I did exactly that then I ended up with x^2 - 2x + (a real number i forgot which)
    Offline

    1
    ReputationRep:
    yes youl need to do what mr m said to find a and b
    then do the long division.

    x^2 - 2x + 5 i believe
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by superfoggy)
    yes youl need to do what mr m said to find a and b
    then do the long division.

    x^2 - 2x + 5 i believe
    So x^2 - 2x + 5 is the real root?

    How do I find the value of A and B?

    Thanks
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by superfoggy)
    yes youl need to do what mr m said to find a and b
    then do the long division.

    x^2 - 2x + 5 i believe
    So x^2 - 2x + 5 is the real root?

    How do I find the value of A and B?

    Thanks
    • Community Assistant
    Offline

    20
    ReputationRep:
    Community Assistant
    I did tell you one method but you ignored it. A quadratic expression cannot be a real root.
    • Thread Starter
    Offline

    0
    ReputationRep:
    please help i am really confused?

    do i need to factorise?
    • Community Assistant
    Offline

    20
    ReputationRep:
    Community Assistant
    Having got this far do this:

    (x - k)(x^2 - 2x + 5) = x^3 + Ax^2 + Bx + 15

    Expand the brackets, collect like terms and equate coefficients (match up each term on the left hand side and the right hand side).
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by Mr M)
    Having got this far do this:

    (x - k)(x^2 - 2x + 5) = x^3 + Ax^2 + Bx + 15

    Expand the brackets, collect like terms and equate coefficients (match up each term on the left hand side and the right hand side).
    ahh i swear to god i did exactly this in the test

    i forgot the = 0 at the end of the first equation
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by Mr M)
    Having got this far do this:

    (x - k)(x^2 - 2x + 5) = x^3 + Ax^2 + Bx + 15

    Expand the brackets, collect like terms and equate coefficients (match up each term on the left hand side and the right hand side).
    (x - k)(x^2 - 2x + 5) = x^3 + Ax^2 + Bx + 15

    x^3 - 2x^2 + 5x - kx^2 + 2kx - 5k = x^3 + Ax^2 + Bx + 15 = 0
    • Thread Starter
    Offline

    0
    ReputationRep:
    can someone tell me what to do from here?
    • Community Assistant
    Offline

    20
    ReputationRep:
    Community Assistant
    (Original post by YouWillNotKnow)
    can someone tell me what to do from here?
    Collect like terms? Look at the numbers on the end and figure out the value of k?
    Offline

    2
    ReputationRep:
    imo, the simplest way to do these is if you have a POLYNOMIAL x^3 + ax^2 +bx +c =0, with roots d, e and f, then c= -def, b=de+ef+df, and a= -(d+e+f)
    • Community Assistant
    Offline

    20
    ReputationRep:
    Community Assistant
    (Original post by Mc^3)
    imo, the simplest way to do these is if you have a quadratic x^3 + ax^2 +bx +c =0, with roots d, e and f, then c= -def, b=de+ef+df, and a= -(d+e+f)
    Are you sure?
    Offline

    2
    ReputationRep:
    (Original post by Mr M)
    Are you sure?
    polynomial, you know what i mean. i think the method's right...
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by Mr M)
    Collect like terms? Look at the numbers on the end and figure out the value of k?
    there are no like terms though :/
    • Community Assistant
    Offline

    20
    ReputationRep:
    Community Assistant
    (Original post by YouWillNotKnow)
    x^3 - 2x^2 + 5x - kx^2 + 2kx - 5k= x^3 + Ax^2 + Bx + 15
    Look at the ends.

    -5k = ?

    So what is k?

    You say there are no like terms but I can see x^2 and x terms ...

    You are struggling badly here. I think you may need to see your teacher for some help.
 
 
 
Poll
Do you agree with the PM's proposal to cut tuition fees for some courses?
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.