16. \( x^{2}+3 x-m-2=0, \operatorname{C} K(S S)=\left\{x_{1}, x_{2}\right\} \) \( 2 x_{1}+x_{2}=-4 \Rightarrow m=? \)

To solve for \( m \) in the equation \( x^2 + 3x - m - 2 = 0 \) given the condition \( 2x_1 + x_2 = -4 \), we start by using Vieta's formulas. According to Vieta's, for a quadratic equation of the form \( ax^2 + bx + c = 0 \), the sum and product of the roots \( x_1 \) and \( x_2 \) are given by: \[ x_1 + x_2 = -\frac{b}{a} \quad \text{and} \quad x_1 x_2 = \frac{c}{a} \] In our case, we have: - \( a = 1 \) - \( b = 3 \) - \( c = -m - 2 \) Thus, we can write: \[ x_1 + x_2 = -\frac{3}{1} = -3 \] \[ x_1 x_2 = \frac{-m - 2}{1} = -m - 2 \] Now, we also have the condition \( 2x_1 + x_2 = -4 \). We can express \( x_2 \) in terms of \( x_1 \): \[ x_2 = -4 - 2x_1 \] Next, we substitute \( x_2 \) into the equation for the sum of the roots: \[ x_1 + (-4 - 2x_1) = -3 \] Simplifying this gives: \[ x_1 - 4 - 2x_1 = -3 \] \[ -x_1 - 4 = -3 \] \[ -x_1 = 1 \quad \Rightarrow \quad x_1 = -1 \] Now, substituting \( x_1 = -1 \) back into the equation for \( x_2 \): \[ x_2 = -4 - 2(-1) = -4 + 2 = -2 \] Now we have the roots \( x_1 = -1 \) and \( x_2 = -2 \). We can now find \( m \) using the product of the roots: \[ x_1 x_2 = -m - 2 \] \[ (-1)(-2) = -m - 2 \] \[ 2 = -m - 2 \] Solving for \( m \): \[ 2 + 2 = -m \] \[ 4 = -m \quad \Rightarrow \quad m = -4 \] Thus, the value of \( m \) is \[ \boxed{-4} \]