Last Activity: 11 Years ago
Ans:p,q
Solution: (x-p)(x-q)-c=0 has roots a and b=> ab=pq-c & a+b=p+q;
similarly if s and t be the roots of (x-a)(x-b)=-c=>st=ab+c & s+t=a+b;(now wkt ab=pq-c=>st=pq)
therfore, s and t are p and q.
the answer is p and -q since in the first equation on equating we get a=c+b and b=c+q when we equate we get a-b=p-q which imples root of equaton 2 is p and -q
(x-p)(x-q)=c
x2-(p+q)x-c=0
hence, a+b=p+q
and a.b=pq-c
a.b+c=p.q
by solving other eq we get
x2-(a+b)x+ab+c=0
thus sum of roots=a+b=p+q
product of roots=a.b+c=p.q
hence its roots are pand q
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