Problem Statement
The cubic equation $2x^3 - 6x^2 + kx - 4 = 0$ has roots $\alpha, \beta, \gamma$.
**(a)** [2 marks] State the values of $\alpha + \beta + \gamma$ and $\alpha\beta\gamma$.
**(b)** [4 marks] Given that the three roots form an arithmetic progression, find the three roots and hence determine the value of $k$.
Verified Solution & Marking Scheme
Part (a) [A1]
$\alpha + \beta + \gamma = -\frac{-6}{2} = 3$
Part (a) [A1]
$\alpha\beta\gamma = -\frac{-4}{2} = 2$
Part (b) [M1]
Roots in AP: Let roots be $a - d, a, a + d$. Sum $= 3a = 3 \implies a = 1$. So one root is $\beta = 1$.
Part (b) [A1]
Substitute $x = 1$ into cubic: $2(1)^3 - 6(1)^2 + k(1) - 4 = 0 \implies 2 - 6 + k - 4 = 0$
Part (b) [A1]
$k - 8 = 0 \implies k = 8$
Part (b) [A1]
Roots: $(1 - d)(1)(1 + d) = 2 \implies 1 - d^2 = 2 \implies d^2 = -1 \implies d = \pm i$. Roots are $1, 1+i, 1-i$.