The roots \(\alpha\) and \(\beta\) of a quadratic equation satisfy \[\alpha + \beta = 4 \qquad \text{and} \qquad \alpha^3 + \beta^3 = 28.\] Using the identi...

A quadratic is completely determined by the sum and the product of its roots, so any question that hands you information about symmetric expressions in \(\alpha\) and \(\beta\) is really asking you to extract those two numbers. Here the sum is given directly and the cube sum is given, and the supplied identity is the bridge between the cube sum and the product.

(a) Show that \(\alpha\beta = 3\) [3]

The identity is printed in the question, so quoting it earns nothing on its own; the marks are for using it. Substitute \(\alpha + \beta = 4\) and \(\alpha^3 + \beta^3 = 28\):

\[28 = 4^3 - 3\alpha\beta(4) \qquad \textbf{[M1]}\]

The [M1] is for the substitution into the identity, and it survives an arithmetic slip afterwards. Evaluating the cube and the multiplier:

\[28 = 64 - 12\alpha\beta \qquad \textbf{[A1]}\]

The first [A1] is for this correct intermediate line. Two slips recur: writing \(4^3 = 12\) instead of \(64\), and computing \(3 \times 4 = 7\) rather than \(12\). Rearranging:

\[12\alpha\beta = 64 - 28 = 36 \quad\Rightarrow\quad \alpha\beta = 3 \qquad \textbf{[A1]}\]

The final [A1] is for the printed target reached with the working shown. Note the direction of the rearrangement: the term \(-12\alpha\beta\) is negative, so moving it across makes it positive while \(28\) crosses in the other direction. Getting that back to front gives \(\alpha\beta = -3\) and loses the mark.

(b) Find a quadratic equation with roots \(\alpha\) and \(\beta\) [2]

A quadratic with sum of roots \(S\) and product of roots \(P\) is

\[x^2 - Sx + P = 0 \qquad \textbf{[M1]}\]

The [M1] is for quoting this standard form, and the minus sign in front of \(S\) is the whole of the difficulty: the sum enters negated, the product does not. Substituting \(S = 4\) and \(P = 3\):

\[x^2 - 4x + 3 = 0 \qquad \textbf{[A1]}\]

The [A1] is for the equation with integer coefficients, as demanded. It must be written as an equation, with "\(= 0\)" present; offering the expression \(x^2 - 4x + 3\) alone does not answer the question asked.

Check: this quadratic factorises as \((x - 1)(x - 3) = 0\), so the roots are \(1\) and \(3\). Their sum is \(1 + 3 = 4\), as given, and their cube sum is \(1^3 + 3^3 = 1 + 27 = 28\), also as given. Both original conditions are satisfied, so the answer is confirmed. That the roots turn out to be pleasant integers is a useful reassurance, but the method was designed to work without ever finding them, which is the point of the technique.