The roots of the equation \(2x^{2} + kx + 5 = 0\) are \(\alpha\) and \(\beta\), where k is a constant. If \(\alpha^{2} + \beta^{2} = -1\), find the values of k.
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Correct Answer: Option C
Explanation:
\(2x^{2} + kx + 5 = 0\)
\(\alpha + \beta = \frac{-b}{a} = \frac{-k}{2}\)
\(\alpha \beta = \frac{c}{a} = \frac{5}{2}\)
\(\alpha^{2} + \beta^{2} = (\alpha + \beta)^{2} - 2\alpha \beta\)
\(-1 = (\frac{-k}{2})^{2} - 2(\frac{5}{2})\)
\(-1 = \frac{k^{2}}{4} - 5 \implies \frac{k^{2}}{4} = 4\)
\(k^{2} = 16 \therefore k = \pm 4\)
\(2x^{2} + kx + 5 = 0\)
\(\alpha + \beta = \frac{-b}{a} = \frac{-k}{2}\)
\(\alpha \beta = \frac{c}{a} = \frac{5}{2}\)
\(\alpha^{2} + \beta^{2} = (\alpha + \beta)^{2} - 2\alpha \beta\)
\(-1 = (\frac{-k}{2})^{2} - 2(\frac{5}{2})\)
\(-1 = \frac{k^{2}}{4} - 5 \implies \frac{k^{2}}{4} = 4\)
\(k^{2} = 16 \therefore k = \pm 4\)