\[ 3 x-y+2=0 \text { and } y=-x^{2}+2 x+8 \] Show that the roots of \( 3 x^{2}+(k+2) x=1-k \) values of k .

To show that the roots of the quadratic equation \( 3x^{2} + (k + 2)x = 1 - k \) can be connected with the lines given by the equations \( 3x - y + 2 = 0 \) and \( y = -x^{2} + 2x + 8 \), we start by substituting \( y \) from the first equation into the second to find the intersection points. First, rewrite the first equation to express \( y \): \[ y = 3x + 2 \] Now substitute this into the second equation: \[ 3x + 2 = -x^{2} + 2x + 8 \] Rearranging gives: \[ x^{2} + x - 6 = 0 \] Factoring this quadratic, we find: \[ (x - 2)(x + 3) = 0 \] Thus, the roots are \( x = 2 \) and \( x = -3 \). To find the corresponding \( y \) values, substitute back into \( y = 3x + 2 \): For \( x = 2 \): \[ y = 3(2) + 2 = 8 \] For \( x = -3 \): \[ y = 3(-3) + 2 = -7 \] Next, these intersection points provide insight into the value of \( k \) in \( 3x^{2} + (k + 2)x = 1 - k \). For this equation to have roots at \( x = 2 \) and \( x = -3 \), we substitute these values into the equation. Let's substitute \( x = 2 \): \[ 3(2)^{2} + (k + 2)(2) = 1 - k \] Calculating, \[ 12 + 2k + 4 = 1 - k \] Which simplifies to: \[ 2k + 16 = 1 - k \] Bringing all \( k \) terms together, we get: \[ 3k = -15 \implies k = -5 \] Now let's check with the second root \( x = -3 \): \[ 3(-3)^{2} + (k + 2)(-3) = 1 - k \] Calculating gives: \[ 27 - 3k - 6 = 1 - k \] Which simplifies to: \[ 21 - 3k = 1 - k \] Arranging yields: \[ 21 - 1 = 3k - k \implies 20 = 2k \implies k = 10 \] Thus, the values of \( k \) where the quadratic has roots at the points of intersection are key links between the geometric aspects of the equations and algebraic formulation of \( k \).