The equation \[3x^2 - 8x + c = 0,\] where \(c\) is a constant, has roots \(\alpha\) and \(\beta\). One root is three times the other, so that \(\beta = 3\al...

The extra piece of information here is a relationship between the roots, \(\beta = 3\alpha\). Combined with the sum of the roots taken from the coefficients, it pins down both roots without any need to solve the quadratic, and the product then delivers the unknown constant \(c\). For \(ax^2 + bx + c = 0\) the standard results are \(\alpha + \beta = -\dfrac{b}{a}\) and \(\alpha\beta = \dfrac{c}{a}\), and the division by \(a\) matters here because the leading coefficient is \(3\), not \(1\).

(a) Find \(\alpha\) and \(\beta\) [3]

From \(3x^2 - 8x + c = 0\), so \(a = 3\) and \(b = -8\):

\[\alpha + \beta = -\frac{-8}{3} = \frac{8}{3} \qquad \textbf{[B1]}\]

The [B1] is for the correct sum, and the two things to get right are the sign, which turns \(-8\) into \(+\tfrac{8}{3}\), and the division by \(3\). Writing \(\alpha + \beta = 8\) is the standard mistake when the leading coefficient is not \(1\).

Now substitute the given relationship \(\beta = 3\alpha\):

\[\alpha + 3\alpha = 4\alpha = \frac{8}{3} \qquad \textbf{[M1]}\] \[\alpha = \frac{8}{12} = \frac{2}{3}, \qquad \beta = 3\alpha = 2 \qquad \textbf{[A1]}\]

The [M1] is for using \(\beta = 3\alpha\) in the sum; the [A1] needs both roots. Note that the question fixes which root is which, so the pair \(\left(\tfrac{2}{3},\, 2\right)\) in that order is what "the value of \(\alpha\) and the value of \(\beta\)" asks for.

(b) Hence find \(c\) [2]

Use the product of the roots, which is where \(c\) lives:

\[\alpha\beta = \frac{c}{3} \qquad \textbf{[M1]}\] \[\frac{2}{3} \times 2 = \frac{4}{3} = \frac{c}{3} \quad\Rightarrow\quad c = 4 \qquad \textbf{[A1]}\]

The [M1] is for quoting the product relationship, the [A1] for the value. Again the \(a = 3\) is easy to lose: \(\alpha\beta = c\) would give \(c = \tfrac{4}{3}\), which is wrong. An equally acceptable route is substitution: since \(\alpha = \tfrac{2}{3}\) is a root, \(3\left(\tfrac{4}{9}\right) - 8\left(\tfrac{2}{3}\right) + c = 0\), giving \(\tfrac{4}{3} - \tfrac{16}{3} + c = 0\) and \(c = 4\).

(c) Find \(\dfrac{1}{\alpha^2} + \dfrac{1}{\beta^2}\) without further solving [3]

The instruction "without further solving" means the answer must come from the symmetric functions rather than from substituting the numbers found in (a). Combine the two fractions over a common denominator:

\[\frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{\beta^2 + \alpha^2}{\alpha^2\beta^2} = \frac{\alpha^2 + \beta^2}{(\alpha\beta)^2} \qquad \textbf{[M1]}\]

The [M1] is for that restructuring, which is the whole idea: an expression in reciprocals of squares has been converted into two quantities already known.

\[\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = \left(\frac{8}{3}\right)^2 - 2\left(\frac{4}{3}\right) = \frac{64}{9} - \frac{8}{3} = \frac{64}{9} - \frac{24}{9} = \frac{40}{9} \qquad \textbf{[A1]}\] \[(\alpha\beta)^2 = \left(\frac{4}{3}\right)^2 = \frac{16}{9}, \qquad \text{so } \frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{40}{9} \div \frac{16}{9} = \frac{40}{16} = \frac{5}{2} \qquad \textbf{[A1]}\]

Two independent accuracy marks, one for \(\alpha^2 + \beta^2\) and one for the final value. Converting \(\tfrac{8}{3}\) to ninths before subtracting is essential; subtracting \(\tfrac{8}{3}\) from \(\tfrac{64}{9}\) as though the denominators matched gives \(\tfrac{56}{9}\) and loses both marks. Also note that dividing by \(\tfrac{16}{9}\) means multiplying by \(\tfrac{9}{16}\), so the ninths cancel neatly.

The wrong turn this part is built to catch is writing \(\dfrac{1}{\alpha^2} + \dfrac{1}{\beta^2} = \dfrac{1}{\alpha^2 + \beta^2}\). Reciprocals do not add like that, as a single numerical case shows: \(\tfrac{1}{1} + \tfrac{1}{4} = 1.25\), whereas \(\tfrac{1}{1 + 4} = 0.2\).

Check: the roots from (a) can be used as verification even though they must not be used as the method. With \(\alpha = \tfrac{2}{3}\) and \(\beta = 2\), \(\dfrac{1}{\alpha^2} = \dfrac{9}{4}\) and \(\dfrac{1}{\beta^2} = \dfrac{1}{4}\), so the sum is \(\dfrac{10}{4} = \dfrac{5}{2}\), matching. And \(3x^2 - 8x + 4 = (3x - 2)(x - 2)\) has roots \(\tfrac{2}{3}\) and \(2\), confirming both (a) and (b).