Correct option is (C) 5
Given Equation is:
\(2 x^2+p x-3=0\)
One root = -3
let the other root = \(\beta\)
\(\alpha \beta =\frac{c}{a}\)
\(-3 \times \beta =\frac{-3}{2}\)
\(-3 \beta =\frac{-3}{2}\)
\(\beta =\frac{-\not3}{2 \times -\not3}=\frac{1}{2}\)
\(\alpha+\beta =-\frac{b}{a}\)
\(\frac{-3}{1}+\frac{1}{2} =\frac{-p}{2}\)
\(-\frac{6+1}{2} =-\frac{p}{2}\)
\(\not-\frac{5}{\not2} =\not-\frac{p}{\not2}\)
\(\therefore p =5\)
