Asked by  | 22nd Sep, 2008, 07:29: PM

Expert Answer:

from given conditions a+b = -p  and ab = 1  and c+d = -q  and cd = 1

E = (a-c)(b-c)(a+d)(b+d)

=(ab-c(a+b) + c2) (ab+d(a+b)+d2 )


now x2+qx+1=0 has its roots as c and d

so c2+qc+1=0 and d2+qd+1=0

i.e. c2+1=-qc  and d2+1=-qd

so E = (pc-qc)(-pd-qd)=-(p-q)(p+q)cd=-(p2-q2)

Answered by  | 23rd Sep, 2008, 01:15: AM

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