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    (Original post by RDKGames)
    Yes the degree equals the total number of roots; real + imaginary. But as Zacken stated, the coefficients need to be real, which they are in this case and many others.
    Ahh that's cool, I've never been taught that before, thanks
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    (Original post by Uni12345678)
    I have no idea if this is correct or not, but...

    Attachment 567000

    EDIT : sorry just realised the last part is a little wrong , b squared - 4ac = -2 squared - 4(1)(-root10) so should be 4 + 4root10 so still 2 real roots but not 16 + 4root10
    I think where I got stuck was because I only factorised x^4 and 4x^3 and left the rest. Thank you for your help !
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    (Original post by physicsmaths)
    Shut up all of u yeh.
    You stupid retards don't even ****ing know how to solve **** and rearrange.
    Hope u all join me in clearing. Spackas
    Help me solve x+1=0
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    Feel like the intended method here is probs differentiating and finding the turning points, since they occur at very nice x values
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    (Original post by 13 1 20 8 42)
    Feel like the intended method here is probs differentiating and finding the turning points, since they occur at very nice x values
    Oh yehhhh I knew I'd picked a stupid method urghh I felt so clever

    EDIT: actually the turning points aren't necessarily at the same values that the graph intersects x so don't think so...
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    (Original post by Uni12345678)
    Oh yehhhh I knew I'd picked a stupid method urghh I felt so clever

    EDIT: actually the turning points aren't necessarily at the same values that the graph intersects x so don't think so...
    They still tell you how many real roots there are. If all the turning points are below the x-axis for a quartic then there are only 2 roots. If one of them is on the x-axis, there are 3 roots. If 1 is above and 2 are below; there are 4 real roots. If 2 turning points are on the x-axis then there are 2 real roots. If all of them are above, then no real roots. (I think I got all those right?)

    If you know the general shape of a quartic this is straight forward.
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    (Original post by Uni12345678)
    Oh yehhhh I knew I'd picked a stupid method urghh I felt so clever

    EDIT: actually the turning points aren't necessarily at the same values that the graph intersects x so don't think so...
    Yes but if you find the turning points, you can see how many fall below and above the x axis and work out how many roots it has
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    Thank you all for your help on the first question, for 89 Ive looked through all my notes but im afraid i havent a clue how to start. All I know is about using CAST diagrams but i dont think i can apply that here either ...thank you once again
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    (Original post by Uni12345678)
    Oh yehhhh I knew I'd picked a stupid method urghh I felt so clever

    EDIT: actually the turning points aren't necessarily at the same values that the graph intersects x so don't think so...
    Hardly stupid. More clever than the alternative, really.

    The point is that with the turning points you can figure out the number of roots without knowing them specifically.

    Trying to find turning points you get

     \displaystyle 4x^3 - 12x^2 + 8x = 0 so  \displaystyle x = 0 or  \displaystyle x^2 - 3x + 2 = 0 which gives  \displaystyle x = 1 or  \displaystyle x = 2

    Clearly the graph is positive if you go left enough, so to get down to its first turning point at x = 0, where we have the function equalling -10, we have to cross the axis. This point will of course be a minimum, preceding a maximum at x = 1. We put this in and see the value is still negative. So the graph must stay under the axis from x = 0 to x = 1. Then x = 2 will be a minimum, so the graph will also stay under the axis from x = 1 to x =2, and we know that eventually the graph will cross the axis again as x^4 is positive and will eventually override the other terms. So there are two roots.
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    (Original post by Jas1947)
    For 83 I keep getting a repeated root where X=-2 but the answer should be C not A. Please explain ...
    It can only have an even number of real roots btw so that eliminates half of em.
    Then the rest is just TPs.


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    (Original post by Jas1947)
    Thank you all for your help on the first question, for 89 Ive looked through all my notes but im afraid i havent a clue how to start. All I know is about using CAST diagrams but i dont think i can apply that here either ...thank you once again
    For 89 you can use the general solutions for sin and tan where:

    For sine:
    sinx=\alpha
    x=2n\pi+arcsin\alpha
    x=2n\pi+\pi-arcsin\alpha

    For tan:
    tanx=\beta
    x=n\pi+arctan\beta

    where n\in\mathbb{Z}

    Given the range for which x can take, you can deduce the two solutions for sine and tan given their domains, hence find the distances between these solutions.

    That's how I did it anyway, got my answer as \frac{\pi}{6} if anyone wants to confirm (unless you're supposed to add the two? In which case it's \frac{\pi}{3}).
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    (Original post by 13 1 20 8 42)
    Hardly stupid. More clever than the alternative, really.

    The point is that with the turning points you can figure out the number of roots without knowing them specifically.

    Trying to find turning points you get

     \displaystyle 4x^3 - 12x^2 + 8x = 0 so  \displaystyle x = 0 or  \displaystyle x^2 - 3x + 2 = 0 which gives  \displaystyle x = 1 or  \displaystyle x = 2

    Clearly the graph is positive if you go left enough, so to get down to its first turning point at x = 0, where we have the function equalling -10, we have to cross the axis. This point will of course be a minimum, preceding a maximum at x = 1. We put this in and see the value is still negative. So the graph must stay under the axis from x = 0 to x = 1. Then x = 2 will be a minimum, so the graph will also stay under the axis from x = 1 to x =2, and we know that eventually the graph will cross the axis again as x^4 is positive and will eventually override the other terms. So there are two roots.
    Oh that's a nice method I think they probably intended it to be solved this way
 
 
 
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