if the sum of two of the roots of x3+px2+qx+r=0 then pq=?
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Answer:
Given,
x
3
−px
2
+qx−r=0
say the cubic equation has roots a,−a,b
then
a+−a+b=(−1)(−p)⟶(1)
a(−a)+a(b)+(−a)b=+q⟶(2)
a(−a)b=(−1)(−r)⟶(3)
∴ From (1) we have,
b=p⟶(4)
From (2) we have
−a
2
=q⟶(5)
From (3) we have
−a
2
b=r⟶(6)
Using (4) & (5) we can write (6) as
+qp=r (∵ −a
2
=q,b=p)
⇒r=pq
Step-by-step explanation:
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