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    (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.


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    (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
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    (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 .
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    (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....
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    (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}}
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    Sorry I was out.

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    (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.


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    (Original post by timebent)
    x
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    (Original post by Chittesh14)
    Sorry I was out.

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    (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.


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    Thank you Thank you Thank you so much ^-^
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    Uh... \sqrt{16} = 4 \neq \pm 4i.
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    (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.
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    (Original post by Zacken)
    Uh... \sqrt{16} = 4 \neq \pm 4i.
    You know what I meant... -16 Zacky boy.


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


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    (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.


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    yup...
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    (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.
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    (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
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    (Original post by timebent)
    x
    Did u manage to solve it in the end?


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


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


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