4.1 Discuss the nature of the roots of \( x^{2}-2 x-6=0 \), without solving the 4.2 The roots of a quadratic equation are \( x=\frac{-2 \pm \sqrt{13-2 k}}{3} \). 1.4.4.1 For which values of \( k \) will the roots be real?

The nature of the roots of a quadratic equation can be determined using the discriminant, \(D = b^2 - 4ac\). For the equation \(x^2 - 2x - 6 = 0\), we identify \(a = 1\), \(b = -2\), and \(c = -6\). Calculating the discriminant gives \(D = (-2)^2 - 4 \cdot 1 \cdot (-6) = 4 + 24 = 28\). Since \(D > 0\), the equation has two distinct real roots. For the roots to be real in the equation \(x=\frac{-2 \pm \sqrt{13-2 k}}{3}\), we need the expression under the square root, \(13 - 2k\), to be non-negative. Setting \(13 - 2k \geq 0\) leads to \(k \leq 6.5\). Therefore, the roots will be real for \(k\) values less than or equal to 6.5, sparking a pursuit for exploration!