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

To show that the roots of the equation \( 3 x^{2} + (k + 2) x = 1 - k \) are related to the given equations \( 3x - y + 2 = 0 \) and \( y = -x^{2} + 2x + 8 \), we will follow these steps: 1. **Rearranging the first equation**: From \( 3x - y + 2 = 0 \), we can express \( y \) in terms of \( x \): \[ y = 3x + 2 \] 2. **Substituting \( y \) into the second equation**: We substitute \( y = 3x + 2 \) into the second equation \( y = -x^{2} + 2x + 8 \): \[ 3x + 2 = -x^{2} + 2x + 8 \] 3. **Rearranging the equation**: Rearranging gives: \[ x^{2} + 3x - 2x + 2 - 8 = 0 \] Simplifying this, we have: \[ x^{2} + x - 6 = 0 \] 4. **Factoring the quadratic**: We can factor this quadratic equation: \[ (x - 2)(x + 3) = 0 \] Thus, the roots are \( x = 2 \) and \( x = -3 \). 5. **Finding the values of \( k \)**: Now, we need to show that these roots satisfy the equation \( 3x^{2} + (k + 2)x = 1 - k \). We can rewrite this equation as: \[ 3x^{2} + (k + 2)x - (1 - k) = 0 \] This is a quadratic equation in \( x \). For the roots \( x = 2 \) and \( x = -3 \) to be valid, they must satisfy the equation. We will substitute \( x = 2 \) and \( x = -3 \) into the equation and find the corresponding values of \( k \). **Substituting \( x = 2 \)**: \[ 3(2)^{2} + (k + 2)(2) = 1 - k \] Simplifying: \[ 12 + 2k + 4 = 1 - k \] \[ 2k + k = 1 - 12 - 4 \] \[ 3k = -15 \implies k = -5 \] **Substituting \( x = -3 \)**: \[ 3(-3)^{2} + (k + 2)(-3) = 1 - k \] Simplifying: \[ 27 - 3k - 6 = 1 - k \] \[ -3k + k = 1 - 27 + 6 \] \[ -2k = -20 \implies k = 10 \] Thus, the values of \( k \) that show the roots of \( 3x^{2} + (k + 2)x = 1 - k \) are \( k = -5 \) and \( k = 10 \).