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I am currently trying to solve under what condition does the following term will be zero?

   2*a*b*n*c*Q - 3*a*b*n*p*c*Q + a*b*n*p^2*c*Q - 4*a^2*b*n*p*Q^2 - 
   4*a*b^2*n*p*Q^2 + 6*a^2*b*n*p^2*Q^2 + 6*a*b^2*n*p^2*Q^2 - 
   2*a^2*b*n*p^3*Q^2 - 2*a*b^2*n*p^3*Q^2 + a^2*b*c*Q^2 + 
   a*b^2*c*Q^2 - 
   2*a^2*b*p*c*Q^2 - 2*a*b^2*p*c*Q^2==0

The assumption is that

0<a<1 && 0<b<1 && 0<c<1 && 0<n<2 && Q>0.  (Also, all of them are real)


Attempt:

Assuming[0 < a < 1 && 0 < b < 1 && 0 < c < 1 && 0 < n < 2 && Q > 0 && 
  Element[a | b | c | n | Q, Reals], Reduce[
  2*a*b*n*c*Q - 3*a*b*n*p*c*Q + a*b*n*p^2*c*Q - 4*a^2*b*n*p*Q^2 - 
    4*a*b^2*n*p*Q^2 + 6*a^2*b*n*p^2*Q^2 + 6*a*b^2*n*p^2*Q^2 - 
    2*a^2*b*n*p^3*Q^2 - 2*a*b^2*n*p^3*Q^2 + a^2*b*c*Q^2 + 
    a*b^2*c*Q^2 - 2*a^2*b*p*c*Q^2 - 2*a*b^2*p*c*Q^2 == 0]]

It returns,

(Q == 0 && (p == 1 || 
     p == 2)) || ((-c + 2 c p + 4 n p - 6 n p^2 + 2 n p^3) Q != 0 && 
   a == (2 c n - 3 c n p + c n p^2 + b c Q - 2 b c p Q - 4 b n p Q + 
     6 b n p^2 Q - 
     2 b n p^3 Q)/((-c + 2 c p + 4 n p - 6 n p^2 + 
       2 n p^3) Q)) || (p == 1/2 && n == 0 && Q != 0) || (Q == 0 && 
   2 - 3 p + p^2 != 0 && n == 0) || (Q == 0 && 
   n (2 - 3 p + p^2) != 0 && c == 0) || ((p == 1 || p == 2) && 
   c == 0 && Q != 0) || (n == 0 && -1 + 2 p != 0 && c == 0 && 
   2 p Q - 3 p^2 Q + p^3 Q != 0) || (p == 0 && n == 0 && c == 0 && 
   Q != 0) || (Q != 0 && p == 0 && 2 n != 0 && c == 0) || a == 0 || 
 b == 0 || Q == 0

Two questions:

1) Did I do this correctly? 2) How do I interpret the solution returned? Isn't that "||" represents OR? So between each "||", then that is a solution?

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1 Answer 1

up vote 3 down vote accepted

Please check the result by doing the following

   eqn = FullSimplify[
   2*a*b*n*c*Q - 3*a*b*n*p*c*Q + a*b*n*p^2*c*Q - 4*a^2*b*n*p*Q^2 - 
     4*a*b^2*n*p*Q^2 + 6*a^2*b*n*p^2*Q^2 + 6*a*b^2*n*p^2*Q^2 - 
     2*a^2*b*n*p^3*Q^2 - 2*a*b^2*n*p^3*Q^2 + a^2*b*c*Q^2 + 
     a*b^2*c*Q^2 - 2*a^2*b*p*c*Q^2 - 2*a*b^2*p*c*Q^2 == 0] // 
  TraditionalForm

This returns

(*a b Q (Q (a+b) (c (2 p-1)+2 n (p-2) (p-1) p)-c n (p-2) (p-1))==0*)

This will allow you to verify the result you've got

Reduce[a b Q (-c n (-2 + p) (-1 + p) + (a + 
        b) (2 n (-2 + p) (-1 + p) p + c (-1 + 2 p)) Q) == 0]

(Q == 0 && (p == 1 || 
     p == 2)) || ((-c + 2 c p + 4 n p - 6 n p^2 + 2 n p^3) Q != 0 && 
   a == (2 c n - 3 c n p + c n p^2 + b c Q - 2 b c p Q - 4 b n p Q + 
     6 b n p^2 Q - 
     2 b n p^3 Q)/((-c + 2 c p + 4 n p - 6 n p^2 + 
       2 n p^3) Q)) || (p == 1/2 && n == 0 && Q != 0) || (Q == 0 && 
   2 - 3 p + p^2 != 0 && n == 0) || (Q == 0 && 
   n (2 - 3 p + p^2) != 0 && c == 0) || ((p == 1 || p == 2) && 
   c == 0 && Q != 0) || (n == 0 && -1 + 2 p != 0 && c == 0 && 
   2 p Q - 3 p^2 Q + p^3 Q != 0) || (p == 0 && n == 0 && c == 0 && 
   Q != 0) || (Q != 0 && p == 0 && 2 n != 0 && c == 0) || a == 0 || 
 b == 0 || Q == 0

As you stated each proposition between || is a solution. Looking at the last three you see that if a = 0 or b = 0 or Q = 0 the equation is equal to zero.

you can try it out. pick a segment and perform the following for (p == 0 && n == 0 && c == 0 && Q != 0)

eqn/.{p->0,n->0,c->0}

(*True*)
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