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    (Original post by Chittesh14)
    Is my answer right in terms of explanation or have I blundered somewhere.
    The answer is 3 I got that .

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    You say it cant be negative but you don't explain why. Use the cubic for that.


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    (Original post by Chittesh14)
    Is my answer right in terms of explanation or have I blundered somewhere.
    The answer is 3 I got that .

    Attachment 561556
    You've got two simultaneous equations, you've only used one equation. It stands to reason that you need to use the other as well. Indeed, if you check a = \pm 3, in the cubic, you'll find that a=3 is the only valid solution.
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    (Original post by Zacken)
    You've got two simultaneous equations, you've only used one equation. It stands to reason that you need to use the other as well. Indeed, if you check a = \pm 3, in the cubic, you'll find that a=3 is the only valid solution.
    Oh lol. Thanks, yeah even if I put a = -3 in my other simultaneous equation, it doesn't work.


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    (Original post by RDKGames)
    You say it cant be negative but you don't explain why. Use the cubic for that.


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    Thanks, I just used the second simultaneous equation lol it's shorter that way.


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    I tried working out part b by using the argument of Z1 and Z2 and multiplying them Together and dividing them but the answers were always wrong until I actually found Z1Z2 and then worked out the argument. Does is rule only apply for modulus?
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    (Original post by Chittesh14)
    I tried working out part b by using the argument of Z1 and Z2 and multiplying them Together and dividing them but the answers were always wrong until I actually found Z1Z2 and then worked out the argument. Does is rule only apply for modulus?
    It can be easier to represent complex numbers in their exponential form.
     \displaystyle a+bi \equiv re^{ix}, \ r=\sqrt{a^2+b^2}, \ x=\arctan \frac{b}{a} .
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    (Original post by Chittesh14)
    I tried working out part b by using the argument of Z1 and Z2 and multiplying them Together and dividing them but the answers were always wrong until I actually found Z1Z2 and then worked out the argument. Does is rule only apply for modulus?
    No. For the argument you add/subtract the individual arguments respectively rather than multiply/divide. You should find out what really is going on visually when two complex numbers multiply/divide in order to understand it, or just prove it to yourself algebraically.

    I take it you are doing further maths if you're looking at complex numbers?
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    (Original post by B_9710)
    It can be easier to represent complex numbers in their exponential form.
     \displaystyle a+bi \equiv re^{ix}, \ r=\sqrt{a^2+b^2}, \ x=\arctan \frac{b}{a} .
    Sorry, I don't understand what you mean by the re^{ix} part. I understand the r as modulus. I understand x as a part in working out the argument, but I'm not too sure about e and i or whatever because maybe I haven't learnt it? :O.

    (Original post by RDKGames)
    No. For the argument you add/subtract the individual arguments respectively rather than multiply/divide. You should find out what really is going on visually when two complex numbers multiply/divide in order to understand it, or just prove it to yourself algebraically.

    I take it you are doing further maths if you're looking at complex numbers?
    Yeah, I was going to do that lol and actually do it by working out Z1 * Z2 and then doing the argument. That's what I did at the end, but I was too mad by then to actually explore it lol. So, it's like powers / exponents. In terms of the same power (argument), you add or subtract.

    Yeah, FP1 - Complex Numbers lolz.
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    (Original post by Chittesh14)
    Sorry, I don't understand what you mean by the re^{ix} part. I understand the r as modulus. I understand x as a part in working out the argument, but I'm not too sure about e and i or whatever because maybe I haven't learnt it? :O.

    That's in FP2 and is simply a different form of a complex number, don't worry about it.
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    (Original post by RDKGames)
    You say it cant be negative but you don't explain why. Use the cubic for that.


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    (Original post by Zacken)
    You've got two simultaneous equations, you've only used one equation. It stands to reason that you need to use the other as well. Indeed, if you check a = \pm 3, in the cubic, you'll find that a=3 is the only valid solution.
    (Original post by B_9710)
    It can be easier to represent complex numbers in their exponential form.
     \displaystyle a+bi \equiv re^{ix}, \ r=\sqrt{a^2+b^2}, \ x=\arctan \frac{b}{a} .
    Nvm guys, sorry.
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    (Original post by Chittesh14)
    Nvm guys, sorry.
    For c it's simply the case of showing that the angles from that -2 to the other other points add up to 90

    Or you can also do it by finding gradients, they should be negative reciprocal sod each other
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    (Original post by RDKGames)
    For c it's simply the case of showing that the angles from that -2 to the other other points add up to 90

    Or you can also do it by finding gradients, they should be negative reciprocal sod each other
    Lol. I just joined up the sides and it formed a right angled triangle lmfao. That's enough proof!!!
    Yeah, I know that m1m2 = -1 rule for perpendicular lines lol.
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    (Original post by Chittesh14)
    Lol. I just joined up the sides and it formed a right angled triangle lmfao. That's enough proof!!!
    Yeah, I know that m1m2 = -1 rule for perpendicular lines lol.
    "Yeah it looks right-angled enough.... QED!"

    Try that in the exams
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    (Original post by RDKGames)
    "Yeah it looks right-angled enough.... QED!"

    Try that in the exams
    Unfortunately, I might not get full UMS if I try it. So, fk the risk lol.
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    (Original post by RDKGames)
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    Hi guys, I'm just feeling a bit crazy XD I've done FP1 - Chapter 2 - Linear Interpolation. It's taking so long like 20 minutes per question or less idek (it feels long), or more. But, I may be doing something wrong or extra - maybe there is a shorter way of the method idk.

    I'll post my working out to the question later, but I'll post the question now - can any of you just do the question and post your working out too - I might compare it before I post my working out too lol. I don't really know when I can post it lol that's the problem XD.

    Question:

    2. (a) Show that a root of the equation 5x3 - 8x2 + 1 = 0 has a root between x = 1 and x = 2.
    (b) Find this root using linear interpolation correct to one decimal place.
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    (Original post by Chittesh14)
    2. (a) Show that a root of the equation 5x3 - 8x2 + 1 = 0 has a root between x = 1 and x = 2.
    (b) Find this root using linear interpolation correct to one decimal place.
    Eh? This takes less than a minute. Denote your function by f, then since we have f(1) = -2 and f(2) = 9, then since f is continuous, there exists a root \alpha \in (1, 2).

    Now, linearly interpolate f in the above interval, we have

    \displaystyle

\begin{equation*}\frac{2 - \alpha}{9} = \frac{\alpha - 1}{2} \Rightarrow 4 - 2\alpha = 9\alpha - 9 \Rightarrow 11\alpha =  13 \Rightarrow \alpha \approx 1.2 \end{equation*}

    And then interpolate again in (1, 1.2) etc...
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    (Original post by Zacken)
    Eh? This takes less than a minute. Denote your function by f, then since we have f(1) = -2 and f(2) = 9, then since f is continuous, there exists a root \alpha \in (0, 1).

    Now, linearly interpolate f in the above interval, we have

    \displaystyle

\begin{equation*}\frac{2 - \alpha}{9} = \frac{\alpha - 1}{2} \Rightarrow 4 - 2\alpha = 9\alpha - 9 \Rightarrow 11\alpha =  13 \Rightarrow \alpha \approx 1.2 \end{equation*}
    That's not the way I'm taught it in the book. Also, 1.2 is not the correct answer either so idk lol. In the book, you draw out a sketch then use similar triangles to work out an approximation and keep on repeating the process until you get two successive approximations which round to the same answer when rounded to 1 d.p.
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    (Original post by Zacken)
    Eh? This takes less than a minute. Denote your function by f, then since we have f(1) = -2 and f(2) = 9, then since f is continuous, there exists a root \alpha \in (0, 1).

    Now, linearly interpolate f in the above interval, we have

    \displaystyle

\begin{equation*}\frac{2 - \alpha}{9} = \frac{\alpha - 1}{2} \Rightarrow 4 - 2\alpha = 9\alpha - 9 \Rightarrow 11\alpha =  13 \Rightarrow \alpha \approx 1.2 \end{equation*}
    What you've done is found the next approximation to 1 d.p.
    I done the process that my book showed me and then got 1. (18 recurring) and that rounds to 1.2 which is probably how you got that.
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    (Original post by RDKGames)
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    Show that the largest possible root of the equation x^3 - 2x^2 - 3 = 0 lies in the interval [2, 3].

    Anyone? - I can show it lies within the interval but I'm not sure how to answer it.
    Do I just show that when x > 3 or x < 2, f (x) is not 0 ?

    Also, does anyone know if there is a solution bank for the edexcel books FP1 from Keith Pledger?

    Also what is the different between linear interpolation and interval interpolation - this is not even in the book lol.
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    If there were another root of f, x>3 then there would have to be a turning point in the range  (3,\infty ) .
 
 
 
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