1roots are a and a2
a + a2 = -p/3 ...(1)
a3 = 1
so roots are ω and ω2 .. cube roots of unity
we know 1 + ω + ω2 = 0
now, cube eqn (1)
a3 + a6 + 3a4 + 3a5 = - p3/27
2 + 3( a + a2 ) = - p3/27
2 + (3 x -1) = - p3/27
p3 = 27
So, p = 3
For equation, 3x2+px+3=0 ,p>0 ,if one of the root is square of the other 'p' is:
a.1/3
b.1
c.3
d.2/3
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2 Answers
1
66@mirka...
once you get that a=ω and ω2, the relation
a+a2=-p/3 implies that p=3 since we must have 1+a+a2=0.
There is no need of cubing the equation and doing whatever u did.
