\( -3 x^{2}+(k-5) x=-k+2 \) Show that the roots are real and rational for all \( k \in \mathbb{Z} \). Answer: INsTRUCTION: Usually you would need to show all of your working out for a question like this. For the computer to mark your answer, just fill in the final line showing that the roots are real and rational. So the final line that shows that the

To show that the roots of the equation \[ -3x^{2} + (k-5)x = -k + 2 \] are real and rational for all \( k \in \mathbb{Z} \), we can rearrange the equation into standard quadratic form: \[ -3x^{2} + (k-5)x + (k-2) = 0 \] This can be rewritten as: \[ 3x^{2} - (k-5)x - (k-2) = 0 \] Now, we will use the discriminant \( D \) of the quadratic equation \( ax^2 + bx + c = 0 \), which is given by: \[ D = b^2 - 4ac \] In our case, \( a = 3 \), \( b = -(k-5) \), and \( c = -(k-2) \). Thus, we can calculate the discriminant: \[ D = (-(k-5))^2 - 4 \cdot 3 \cdot (-(k-2)) \] Calculating \( D \): \[ D = (k-5)^2 + 12(k-2) \] Expanding this: \[ D = (k^2 - 10k + 25) + (12k - 24) \] \[ D = k^2 + 2k + 1 \] \[ D = (k + 1)^2 \] Since \( (k + 1)^2 \) is a perfect square, it is always non-negative for all \( k \in \mathbb{Z} \). Therefore, the roots are real. Next, to show that the roots are rational, we note that since \( D \) is a perfect square, the roots can be expressed as: \[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{(k-5) \pm (k + 1)}{6} \] Calculating the two possible roots: 1. \( x_1 = \frac{(k-5) + (k + 1)}{6} = \frac{2k - 4}{6} = \frac{k - 2}{3} \) 2. \( x_2 = \frac{(k-5) - (k + 1)}{6} = \frac{-6}{6} = -1 \) Both roots \( x_1 \) and \( x_2 \) are rational for all integer values of \( k \). Thus, we conclude that the roots are real and rational for all \( k \in \mathbb{Z} \).