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    The question is -
    Solve the simultaneous equations p + q + r = 6, (p^2) + (q^2) + (r^2) = 26 and pqr = -12.

    I was able to make the cubic equation but I don't know how to calculate the values of p, q and r.
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    If you've done it right, the 3 roots of the equation are the values of p, q and r.
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    The equation I got was (x^3) - (6x^2) + (5x) - 12 = 0.

    Now I don't know how to calculate its roots? I am supposed to show working.
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    (Original post by Rishabh95)
    The question is -
    Solve the simultaneous equations p + q + r = 6, (p^2) + (q^2) + (r^2) = 26 and pqr = -12.

    I was able to make the cubic equation but I don't know how to calculate the values of p, q and r.
    The roots are all integers so you should find it quite easy if you use the factor theorem.
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    (Original post by Rishabh95)
    The equation I got was (x^3) - (6x^2) + (5x) - 12 = 0.

    Now I don't know how to calculate its roots? I am supposed to show working.
    Your equation is slightly incorrect.
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    Ok, I think I got the equation wrong!

    I tried another question which is -

    Write down the equation whose roots are p, q and r, where p + q + r = 6, qr + rp + pq = 11 and pqr = 6. Hence solve the equation!

    The equation I got was (x^3) - (6x^2) + (11x) - 6 = 0 which I verified from the answers.

    I am guessing as the question says hence, the answer is straightforward, but I don't know how to calculate!
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    (Original post by Get me off the £\?%!^@ computer)
    The roots are all integers so you should find it quite easy if you use the factor theorem.
    Do you mean its guesswork?
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    (Original post by Rishabh95)
    Do you mean its guesswork?
    Well, educated guesswork. But yeah, basically.
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    (Original post by DFranklin)
    Well, educated guesswork. But yeah, basically.
    So what would you guess in the second question?
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    (Original post by Rishabh95)
    So what would you guess in the second question?
    Look at the coefficients. I see a 1,-6,11,-6. Does that suggest anything?
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    Is it that - there is one between 1 and -6 and two between -6 and 11?
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    You can find a solution easily enough through intelligent guesswork which is what it sounds like you have done but you're not satisfied with that - is that right?

    So you looked at pqr=6 and p+q+r=6 and came up with 1, 2 and 3 - that would seem simple enough - yes?
    When the question says "solve" you are being asked to identify the few interesting facts about a curve that may or may not cut either the x or y axis (I'm speaking generally now) so solving means finding where the curve cuts the x-axis, i.e. where y=0. So - if you arrange an ordinary quadratic equation (by way of a simple starter example) such that:-

    y = x^2+x-6

    and factorise such that:-
    y=(x-2)(x+3)

    to find where the curve cuts the x-axis you need find the values of x that make y=0. When you have factorised you have:-

    y=A x B

    and to make y=0 then either A or B must be 0, so with:-
    y=(x-2)(x+3)

    for instance - either x-2=0 or x+3=0 - solving each of these gives you x=2 or x=-3 and these are the so called roots.

    So if the question says the roots are p, q and r and you have intelligently guessed p=1, q=2, r=3 you instantly get the factorized version of your cubic equation:-
    y=(x-1)(x-2)(x-3)
    which multiplies out to give you:-
    (x^3) - (6x^2) + (11x) - 6

    Now - is it that you didn't want to rely on an intelligent guess?
    You have 3 unknowns and 3 equations so it's a matter of substituting a couple of times to eliminated a couple of variables until you are left with an equation that only has numbers and one variable left, e.g.

    rp+qr+pq=11
    r(p+q) + pq=11 -- But p+q+r=6 so p+q=6-r
    r(6-r) +pq=11 -- and pqr=6 so pq=6/r
    r(6-r)+6/r=11 -- nothgin but numbers and 1 variable r.
    so you end up having to try and solve
    6=r^3 -6r^2 +11r -- factorise it

    6 = r (r^2-6r+11)
    So again you have 6 = A x B
    so A and B could be 1 & 6 or 2 & 3 for instance.
    If you look at the quadratic (r^2-6r+11) you can't factorise it as b^2 < 4ac and all this means is that there are no roots - i.e. the x^2 curve does come low enough to intersect with the x-axis. How low does it go? It's easy enough to sketch but a simple differentiation to find the turning point finds the lowest value this quadratic function can have:-

    y = r^2-6r+11
    dy/dx = 2r-6
    So the min value occurs where 2r-6=0 so 2r=6 and r=3.
    When you put r=3 in r^2-6r+11 you find the min value this quadratic function can have is 2 (this is B in our 6=AxB)

    Remembering we said 6 = r (r^2-6r+11), i.e. 6 = A x B
    so A and B could be 1 & 6 or 2 & 3 for instance.
    So r must be 3 to get
    6 = 3 (2) [where 2 is the min value of our quadratic) from 6 = r (r^2-6r+11)

    From then on it's easy to get p and q. I don't know if this is what you were looking for but HTH.
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    x^3 + (alpha+beta+gamma)x^2 + (alphabeta + betaalpha + gammaalpha)x - (alphabetagamma)
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    (Original post by Rishabh95)
    Ok, I think I got the equation wrong!

    I tried another question which is -

    Write down the equation whose roots are p, q and r, where p + q + r = 6, qr + rp + pq = 11 and pqr = 6. Hence solve the equation!

    The equation I got was (x^3) - (6x^2) + (11x) - 6 = 0 which I verified from the answers.

    I am guessing as the question says hence, the answer is straightforward, but I don't know how to calculate!
    x=1 is a solution

    (x-1)(x^2 -5x + 6 )

    (x-1)(x-2)(x-3) hence x=1,2,3

    fairly straightfoward, since you've already done the tricky part finding the equation
 
 
 
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