\(\pm 16\)
\(\pm 8\)
\(\pm 4\)
\(\pm 2\)
Correct answer is C
\(2x^{2} + kx + 5 = 0\)
\(\alpha + \beta = \frac{-b}{a} = \frac{-k}{2}\)
\(\alpha \beta = \frac{c}{a} = \frac{5}{2}\)
\(\alpha^{2} + \beta^{2} = (\alpha + \beta)^{2} - 2\alpha \beta\)
\(-1 = (\frac{-k}{2})^{2} - 2(\frac{5}{2})\)
\(-1 = \frac{k^{2}}{4} - 5 \implies \frac{k^{2}}{4} = 4\)
\(k^{2} = 16 \therefore k = \pm 4\)