If a,b are the roots of x^2+px+q=0 then x^2+(a+b-ab)x-ab(a+b)=0 has roots

• roots= -p,-q
  As for x^2+px+q=0 ---(i), roots are a & b, then a+b=-p, ab=q
  Eqn x^2+(a+b-ab)x-ab(a+b)=0 ---(ii)
  Substituting values of a+b & ab in (ii),
  we get eqn.(ii)as, x^2-(p+q)x+pq=0, so the roots are -p,-q
  
• 11 years agoHelpfull: Yes(1) No(1)
• Options are1) -p,-q
  2) p,-q
  3) a,b
  4) -a,-b
  5) a,-b
  6) p,q
  7) -p,q
  8) -a,b
  9) ap,bq
  10)None of these
• 11 years agoHelpfull: Yes(0) No(1)
• If a and b are roots then a+b=-p and ab=q,So the equation x^2+(a+b-ab)x-ab(a+b)=0 becomes x^2+(-p-q)x+pq=0.Again let it has roots b and c,then b+c=-(-p-q)=p+q and bc=pq,then (b-c)^2=(b+c)^2-4bc=(p+q)^2-4pq=(p-q)^2,so b-c=p-q,sop we get b=p and c=q,So the roots are p and q.
• 11 years agoHelpfull: Yes(0) No(1)
• @Devendra M
  Your answer is -p and -q( x^2-(p+q)x+pq=0, so the roots are -p,-q),that means these must satisfy the equation x^2-(p+q)x+pq=0,but these do not .You can check.But if you take p and q as your roots the equation becomes 0 for both the roots.So the answer is p and q.
• 11 years agoHelpfull: Yes(0) No(0)