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 o...
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.
Answer Details
The sum of the roots of the quadratic equation \(ax^{2} + bx + c = 0\) is given by the formula \(-\frac{b}{a}\), and the product of the roots is given by the formula \(\frac{c}{a}\).
Therefore, for the quadratic equation \(2x^{2} + kx + 5 = 0\), we have:
\[\alpha + \beta = -\frac{k}{2}\]
\[\alpha \beta = \frac{5}{2}\]
Squaring the first equation, we get:
\[(\alpha + \beta)^{2} = \alpha^{2} + 2\alpha\beta + \beta^{2} = \frac{k^{2}}{4}\]
Substituting the given value of \(\alpha^{2} + \beta^{2} = -1\), we have:
\[-1 + 2\alpha\beta = \frac{k^{2}}{4}\]
Substituting the value of \(\alpha \beta = \frac{5}{2}\), we have:
\[-1 + 2\left(\frac{5}{2}\right) = \frac{k^{2}}{4}\]
Simplifying the left-hand side, we get:
\[4 = \frac{k^{2}}{4}\]
Multiplying both sides by 4, we get:
\[16 = k^{2}\]
Taking the square root of both sides, we get:
\[k = \pm 4\]
Therefore, the values of k are \(\pm 4\).