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    (Original post by KloppOClock)
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    first equation rearranged gives  \displaystyle x^3  +ax^2 - bx - c = 0

    differentiating this and setting it equal to 0 to find stationary points;  \displaystyle 3x^2 +2ax - b = 0

    quadratic formula and solving for x gives us two solutions, namely  \displaystyle x_{1} = \frac{-a + \sqrt{a^2 + 3b}}{3} and  \displaystyle x_2 = \frac{-a - \sqrt{a^2 + 3b}}{3} . Now since you are told all the numbers are positive  \displaystyle x_1 > 0 and  \displaystyle x_2 < 0 .

    Differentiating again we get  \displaystyle 6x + 2a , its clear to see that a positive x value will > 0 and therefore minimum and therefore our negative x value is our maximum

    Also since c > 0, and therefore -c < 0, we can sketch the first equation like so;

    http://imgur.com/a/CsPoB

    Do the exact same thing for the second equation and you'll sketch a graph which will have one negative and two positive roots
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    (Original post by DylanJ42)
    first equation rearranged gives  \displaystyle x^3  +ax^2 - bx - c = 0

    differentiating this and setting it equal to 0 to find stationary points;  \displaystyle 3x^2 +2ax - b = 0

    quadratic formula and solving for x gives us two solutions, namely  \displaystyle x_{1} = \frac{-a + \sqrt{a^2 + 3b}}{3} and  \displaystyle x_2 = \frac{-a - \sqrt{a^2 + 3b}}{3} . Now since you are told all the numbers are positive  \displaystyle x_1 &gt; 0 and  \displaystyle x_2 &lt; 0 .

    Differentiating again we get  \displaystyle 6x + 2a , its clear to see that a positive x value will > 0 and therefore minimum and therefore our negative x value is our maximum

    Also since c > 0, and therefore -c < 0, we can sketch the first equation like so;

    http://imgur.com/a/CsPoB

    Do the exact same thing for the second equation and you'll sketch a graph which will have one negative and two positive roots
    Okay so you found the:
    y- intercept and the x values of the max and minimum

    but how do you know if the roots are imaginary or real
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    (Original post by KloppOClock)
    Okay so you found the:
    y- intercept and the x values of the max and minimum

    but how do you know if the roots are imaginary or real
    they cant be imaginary, for that to happen  \displaystyle a^2 + 3b &lt; 0 but you are told that b>0 on the top line
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    (Original post by KloppOClock)
    Okay so you found the:
    y- intercept and the x values of the max and minimum

    but how do you know if the roots are imaginary or real
    I see what you mean now, my bad; (also i labelled the imgur picture wrong, x1 and x2 are mixed up, so ignore the last post and ill fix it up plus add in the extra bit to prove roots are real)

    first equation rearranged gives , im going to call this  \displaystyle f(x) for handiness


    differentiating this and setting it equal to 0 to find stationary points;

    quadratic formula and solving for x gives us two solutions, namely  \displaystyle x_1 = \frac{-a-\sqrt{a^2 + 3b}}{3} . and  \displaystyle x_2 = \frac{-a+\sqrt{a^2 + 3b}}{3}

    Now since you are told all the numbers are positive  \displaystyle x_1 &lt; 0 and  \displaystyle x_2 &gt; 0

    Differentiating again we get , its clear to see that a positive x value will > 0 and therefore minimum and therefore our negative x value is our maximum

    Now the third line tells us that the three roots are real, its clear that since  \displaystyle f(x_2) is the minimum tp its less than -c which is less than 0, we can write  \displaystyle f(x_2) &lt; -c &lt; 0 .

    Now for the third line to be true,  \displaystyle f(x_1) &gt; 0 , so writing this out fully we get;

     \displaystyle (x_1)^3 +a(x_1) ^2 - b(x_1) - c &gt; 0

    so to summarize we know  \displaystyle x_1 &lt; 0 and from line 3 of the question know  \displaystyle f(x_1) &gt; 0 , we also know that the  \displaystyle \text{y intercept -c} &lt; 0 ,  \displaystyle x_2 &gt; 0 and  \displaystyle f(x_2) &lt; 0 so the sketch of f(x) looks like so;

    http://imgur.com/a/3kPfb

    --------
    --------

    Okay now for the second equation,

    ill call this equation  \displaystyle g(x) for handiness

    doing the same thing we get roots  \displaystyle x_3 = \frac{a-\sqrt{a^2 + 3b}}{3} and  \displaystyle x_4 = \frac{a+\sqrt{a^2 + 3b}}{3} . Once again since a,b,c > 0 this means  \displaystyle x_3 &lt; 0 and  \displaystyle x_4 &gt; 0

    Again your happy that  \displaystyle g(x_3) &gt; 0 since its the maximum tp and must be  \displaystyle &gt; c &gt; 0 . We can write this as one inequality as  \displaystyle g(x_3) &gt; c &gt; 0 .

    Now we need to prove that  \displaystyle g(x_4) &lt; 0 , do you notice how  \displaystyle x_4 = -x_1 . So subbing  \displaystyle -x_1 into g(x) yields;

     \displaystyle g(x_4) = g(-x_1) = (-x_1)^3 - a(-x_1)^2 -b(-x_1) + c

     \displaystyle g(x_4) = -(x_1)^3 - a(x_1)^2 +b(x_1) + c

     \displaystyle g(x_4) = -[(x_1)^3 + a(x_1)^2 - b(x_1) - c]

    Remember back to the first equation when line 3 told us that  \displaystyle (x_1)^3 +a(x_1) ^2 - b(x_1) - c &gt; 0  therefore  \displaystyle g(x_4) = -[(x_1)^3 + a(x_1)^2 - b(x_1) - c] &lt; 0 just as we wanted.

    So we know  \displaystyle x_3 &lt; 0 ,  \displaystyle g(x_3) &gt; 0 , the  \displaystyle \text{y intercept c} &gt; 0,  \displaystyle x_4 &gt; 0 and that  \displaystyle g(x_4) &lt; 0 . Sketching this gives us;

    http://imgur.com/a/2NInI

    so  \displaystyle g(x) has a negative and 2 positive real roots, as required

    sorry for this being so long, I didnt realise it involved so much working out (i originally thought this was a MAT multiple choice type question but obviously not)
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    wait till that mauritian priickk comes along and takes this down because it violates the rules smh
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    (Original post by DylanJ42)

    Now we need to prove that  \displaystyle g(x_4) &lt; 0 , do you notice how  \displaystyle x_4 = -x_1 .
    no

    (Original post by fksociety)
    wait till that mauritian priickk comes along and takes this down because it violates the rules smh
    what?
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    (Original post by KloppOClock)
    no
     \displaystyle x_1 = \frac{-a - \sqrt{a^2 + 3b}}{3} and  \displaystyle x_4 = \frac{a + \sqrt{a^2 + 3b}}{3}

    multiplying the left hand side one by -1 will give you the right hand side one

    ie  \displaystyle x_1 \times -1 = x_4

    (Original post by KloppOClock)
    what?
    hes referring to Zacken taking this down for full solution
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    (Original post by DylanJ42)
     \displaystyle x_1 = \frac{-a - \sqrt{a^2 + 3b}}{3} and  \displaystyle x_4 = \frac{a + \sqrt{a^2 + 3b}}{3}

    multiplying the left hand side one by -1 will give you the right hand side one

    ie  \displaystyle x_1 \times -1 = x_4



    hes referring to Zacken taking this down for full solution
    damn i get it now, but that seems like a difficult question to get in under 3 minutes, thanks.

    also, whats wrong with posting the full solution to a question and why would that anger zacken
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    (Original post by KloppOClock)
    damn i get it now, but that seems like a difficult question to get in under 3 minutes, thanks.

    also, whats wrong with posting the full solution to a question and why would that anger zacken
    yea there's probably a much easier way to do this, but that's my best attempt

    its against forum rules to post full solutions and since Zacken got modded or whatever he had to tell people off for posting full solutions as they were breaking the rules. Others tell people off for it too but Zacken seems to be a favourite with the trolls so they pick on him :dontknow:
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    (Original post by KloppOClock)
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    just realised how to do this much much quicker

    call the first equation \displaystyle  f(x) and say  \displaystyle  f(n) = n^3 + an^2 - bn - c  = m for some value of n

    call the second equation  \displaystyle g(x), now  \displaystyle g(-n) = -n^3 -an^2 +bn + c = -[n^3 + an^2 - bn - c] = -m

    so  \displaystyle g(-x) = -f(x)

    therefore g(x) is just f(x) reflected the the x axis and the y axis (or vice versa)

    do some sketches and get; http://imgur.com/a/LqM0S
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    (Original post by DylanJ42)
    just realised how to do this much much quicker

    call the first equation \displaystyle  f(x) and say  \displaystyle  f(n) = n^3 + an^2 - bn - c  = m for some value of n

    call the second equation  \displaystyle g(x), now  \displaystyle g(-n) = -n^3 -an^2 +bn + c = -[n^3 + an^2 - bn - c] = -m

    so  \displaystyle g(-x) = -f(x)

    therefore g(x) is just f(x) reflected the the x axis and the y axis (or vice versa)

    do some sketches and get; http://imgur.com/a/LqM0S
    but how would you know to sub -n into g(x)
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    (Original post by KloppOClock)
    but how would you know to sub -n into g(x)
    because of the differences in signs between the individual terms in f(x) and g(x).
 
 
 
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