a(x-x1)(x-x2)=ax^2-a(x1+x2)x+ax1x2=ax^2+bx+c=0. b=-a(x1+x2), c=ax1x2 and b^2-4ac=a^2(x1+x2)^2-4a^2x1x2=a^2(x1-x2)^2.
Similarly, q=-p(x3+x4), c=px3x4 and p^2-4pr=p^2(x3-x4)^2, so (b^2-4ac)/(q^2-4pr)=(a^2(x1-x2)^2)/(p^2(x3-x4)^2) or ((a(x1-x2))/(p(x3-x4)))^2, where x3 and x4 are roots of the second polynomial. If x3 and x4 are replaced by x^3 and x^4 then the division becomes ((a(x1-x2))/(p(x^3-x^4)))^2.
(i)
If AP is x1, x2, 1/x3, 1/x4, then if d is the common difference, x2=x1+d so x1-x2=-d or (x1-x2)^2=d^2. If 1/x4-1x3=d, then x3-x4=dx3x4, x4(1+dx3)=x3, x4=x3/(1+dx3). Also, if 1/x3-x2=d, then 1-x2x3=dx3 and x3(d+x2)=1, x3=1/(d+x2). Let A=x1 then x2=A+d, 1/x3=A+2d, 1/x4=A+3d. x3=1/(A+2d), x4=1/(A+3d). x3-x4=d/((A+3d)(A+2d)). The division becomes: (a(A+2d)(A+3d)/p)^2=(a(x1+2d)(x1+3d)/p)^2.
(ii)
If a, b, c, form a GP, then b=ar, c=ar^2, where r=common ratio. b^2-4ac=a^2r^2-4a^2r^2=-3a^2r^2. Since square root of b^2-4ac must be greater than or equal to zero for real roots of the quadratic to exist, if a, b and c are consecutive terms in a GP, then there can be no real roots. The imaginary part of the root is given by +arisqrt(3)/2a=+risqrt(3)/2 where i is the imaginary square root of -1, and x1 and x2 are both complex numbers.
If x1, x2, x3, x4 form a GP with common ratio R, x1=A, x2=AR, x3=AR^2, x4=AR^3.
q=-p(x3+x4)=-pAR^2(1+R) and r=px3x4=pA^2R^5. p, q, r form a series p, -pAR^2(1+R), pA^2R^5, which is not an obvious GP, because there is no visible common ratio. However, if we call the supposed common ration X, then R must be such that pX=-pAR^2(1+R) and pX^2=pA^2R^5, making X=-pA^2R^5/pAR^2(1+R)=-AR^3/(1+R) and X=-AR^2(1+R). So R^3/(1+R)=R^2(1+R); R=(1+R)^2; R^2+R+1=0, which has no real roots. The solution is a complex number: (-1+isqrt(3))/2 or (-1-isqrt(3))/2.
From this, X=-A(1-3+2isqrt(3))/2isqrt(3)=x1(1+isqrt(3))/isqrt(3)=-x1(isqrt(3)/3+1). This common ratio applies only to p, q and r.
