Turn on thread page Beta
    Offline

    19
    ReputationRep:
    (Original post by timebent)
    oh ffs everytime why doesn't it go in *rips eyeballs out

    http://www.examsolutions.net/a-level...e/paper.php#Q3
    is k 22? no it's not i got it now

    no can you do part b for me pls i don't even know what i'm doing -.-'

    k=30 though
    As the other user said, the way to go is algebraic division. I used (x - 1/2) as a factor but the other form is also applicable if you don't like to deal with fractions.


    Posted from TSR Mobile
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by B_9710)
    If x=1/2 is a solution then (2x-1) is a factor of the cubic.
    Algebraic division is the way forward.
    that's what i did but it doesn't work eventually i just get x^2 -4x+2 which doesn't give be a complex conjugate, it just gives me root 2 +2

    so you'll be asking for my workings
    Attached Images
     
    Offline

    15
    ReputationRep:
    (Original post by timebent)
    that's what i did but it doesn't work eventually i just get x^2 -4x+2 which doesn't give be a complex conjugate, it just gives me root 2 +2

    so you'll be asking for my workings
    Your very last step when doing algebraic division is wrong. You should get the quotient to be  x^2-4x+13 .
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by B_9710)
    Your very last step when doing algebraic division is wrong. You should get the quotient to be  x^2-4x+13 .
    oh -.-' everytime....
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by B_9710)
    Your very last step when doing algebraic division is wrong. You should get the quotient to be  x^2-4x+13 .
    http://www.examsolutions.net/a-level...y/paper.php#Q5
    i decided to start with the left bracket.

    So i tried to solve for x and see what i'd get, since the question told me there'll be 2 conjugate pairs in total i should get 1 pair from this bracket on the left.

    4x^2 +9=0



4x^2=-9



x^2=-\frac{9}{4}



x=\sqrt {-\frac{9}{4}}
    Offline

    19
    ReputationRep:
    Sorry I was out.

    Name:  ImageUploadedByStudent Room1469211240.697405.jpg
Views: 48
Size:  145.5 KB


    Posted from TSR Mobile
    Offline

    19
    ReputationRep:
    (Original post by timebent)
    http://www.examsolutions.net/a-level...y/paper.php#Q5
    i decided to start with the left bracket.

    So i tried to solve for x and see what i'd get, since the question told me there'll be 2 conjugate pairs in total i should get 1 pair from this bracket on the left.

    4x^2 +9=0



4x^2=-9



x^2=-\frac{9}{4}



x=\sqrt {-\frac{9}{4}}
    Basically, anything which is negative and is square rooted.
    It is basically, that number square rooted i.

    I'll post examples wait.


    Posted from TSR Mobile
    Offline

    19
    ReputationRep:
    (Original post by timebent)
    x
    Name:  ImageUploadedByStudent Room1469211652.715922.jpg
Views: 50
Size:  84.1 KB


    Posted from TSR Mobile
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by Chittesh14)
    Sorry I was out.

    Name:  ImageUploadedByStudent Room1469211240.697405.jpg
Views: 48
Size:  145.5 KB


    Posted from TSR Mobile
    (Original post by Chittesh14)
    Basically, anything which is negative and is square rooted.
    It is basically, that number square rooted i.

    I'll post examples wait.


    Posted from TSR Mobile
    Thank you Thank you Thank you so much ^-^
    Offline

    22
    ReputationRep:
    Uh... \sqrt{16} = 4 \neq \pm 4i.
    Offline

    22
    ReputationRep:
    (Original post by timebent)
    http://www.examsolutions.net/a-level...y/paper.php#Q5
    i decided to start with the left bracket.

    So i tried to solve for x and see what i'd get, since the question told me there'll be 2 conjugate pairs in total i should get 1 pair from this bracket on the left.

    4x^2 +9=0



4x^2=-9



x^2=-\frac{9}{4}



x=\sqrt {-\frac{9}{4}}
    Your last step is wrong, x^2 = a \iff x = \pm \sqrt{a} so x = \pm \frac{3}{2} i which are indeed conjugate pairs.
    Offline

    19
    ReputationRep:
    (Original post by Zacken)
    Uh... \sqrt{16} = 4 \neq \pm 4i.
    You know what I meant... -16 Zacky boy.


    Posted from TSR Mobile
    Offline

    19
    ReputationRep:
    (Original post by timebent)
    Thank you Thank you Thank you so much ^-^
    Np. Try to follow what Zacken said and solve the second bracket.


    Posted from TSR Mobile
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by Zacken)
    Your last step is wrong, x^2 = a \iff x = \pm \sqrt{a} so x = \pm \frac{3}{2} i which are indeed conjugate pairs.
    oh oops i mean to put 3/2 there not the 9 and 4
    thanks ^-^
    (Original post by Chittesh14)
    Np. Try to follow what Zacken said and solve the second bracket.


    Posted from TSR Mobile
    yup...
    Offline

    22
    ReputationRep:
    (Original post by timebent)
    oh oops i mean to put 3/2 there not the 9 and 4
    thanks ^-^
    Uh, no, that wasn't the mistake... the mistake was not including the \pm.
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by Zacken)
    Uh, no, that wasn't the mistake... the mistake was not including the \pm.
    Oh ;_; when will i stop making these small mistakes
    Offline

    19
    ReputationRep:
    (Original post by timebent)
    x
    Did u manage to solve it in the end?


    Posted from TSR Mobile
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by Zacken)
    Uh, no, that wasn't the mistake... the mistake was not including the \pm.
    not sure if the method i used is called factor theorem or inspection
    http://www.examsolutions.net/a-level...y/paper.php#Q5
    but for this question if it gives me one complex number and there's 3 roots then i know it's got to be a pair

    and i used inspection or factor theorem or whatever it's called and tried 2 and got lucky so where would i get my marks assuming that this is my working

    z_2 =3-i because it has a pair

    let z^3 -8z^2 +22z-20=f(z)

    f(2)=8-32+44-20=0

    therefore z=2 thus z-2 is a factor
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by Chittesh14)
    Did u manage to solve it in the end?


    Posted from TSR Mobile
    yes i got the pair for the other bracket easily and i drew the diagram correctly
    Offline

    19
    ReputationRep:
    (Original post by timebent)
    yes i got the pair for the other bracket easily and i drew the diagram correctly
    Great .


    Posted from TSR Mobile
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: July 28, 2016

2,254

students online now

800,000+

Exam discussions

Find your exam discussion here

Poll
Should predicted grades be removed from the uni application process
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.