Vieta gives $a+b+c=3/2$. Let $a+b=1$; then $c=1/2$. Also $abc=1/2$,so $ab=1$. The pairwise product sum is $ab+c(a+b)=1+1/2=3/2=p/2$. Hence **p=3,C**. The algebra permits complex roots; no assumption of three real roots is needed.
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