The equation \(2x^2 - 5x + 4 = 0\) has roots \(\alpha\) and \(\beta\). Write down the value of \(\alpha + \beta\) and the value of \(\alpha\beta\). (2) With...

This question is about symmetric functions of the roots of a quadratic. The instruction "Without solving the equation" in part (b) is the key: the roots of \(2x^{2} - 5x + 4 = 0\) are not real, since the discriminant is \(25 - 32 = -7\), so any attempt to find them individually would fail. Yet the sum and the product of the roots are perfectly ordinary numbers, and every symmetric expression in the roots can be built from those two.

(a) The sum and product of the roots [2 marks]

For \(ax^{2} + bx + c = 0\) with roots \(\alpha\) and \(\beta\):

\[\alpha + \beta = -\frac{b}{a}, \qquad \alpha\beta = \frac{c}{a}.\]

Here \(a = 2\), \(b = -5\) and \(c = 4\), so

\[\alpha + \beta = -\frac{-5}{2} = \frac{5}{2} \qquad \textbf{B1}\] \[\alpha\beta = \frac{4}{2} = 2 \qquad \textbf{B1}\]

These are independent B marks, one each, so a slip in one does not cost the other. The sign is the trap in the sum: the formula carries a minus, and \(b\) is itself negative, so the two negatives give a positive \(\dfrac{5}{2}\). Writing \(-\dfrac{5}{2}\) is the commonest error, and it propagates through both remaining parts. The other frequent slip is forgetting to divide by \(a\), giving \(5\) and \(4\) instead.

(b) The value of \(\alpha^{2} + \beta^{2}\) [2 marks]

Squaring the sum produces the wanted squares plus an unwanted cross term:

\[\left(\alpha + \beta\right)^{2} = \alpha^{2} + 2\alpha\beta + \beta^{2},\]

so subtracting twice the product isolates what is needed. This identity is the whole method:

\[\alpha^{2} + \beta^{2} = \left(\alpha + \beta\right)^{2} - 2\alpha\beta \qquad \textbf{M1}\]

The M1 is for quoting a correct identity and stands even if the substituted values are wrong, which is why the identity should be written before the numbers go in. Substituting:

\[\alpha^{2} + \beta^{2} = \left(\frac{5}{2}\right)^{2} - 2\left(2\right) = \frac{25}{4} - 4 = \frac{25 - 16}{4} = \frac{9}{4} \qquad \textbf{A1}\]

The wrong turn here is writing \(\alpha^{2} + \beta^{2} = \left(\alpha + \beta\right)^{2}\), that is \(\dfrac{25}{4}\), which ignores the cross term entirely. A second error is subtracting \(\alpha\beta\) rather than \(2\alpha\beta\). Notice that the answer is positive even though the roots themselves are not real; a sum of squares of complex conjugates can be positive, negative or zero, so a positive value here is no cause for alarm.

(c) The value of \(\alpha^{3} + \beta^{3}\) [2 marks]

The same strategy applies one degree higher. Cubing the sum gives

\[\left(\alpha + \beta\right)^{3} = \alpha^{3} + 3\alpha^{2}\beta + 3\alpha\beta^{2} + \beta^{3} = \alpha^{3} + \beta^{3} + 3\alpha\beta\left(\alpha + \beta\right),\]

since the two middle terms share the factor \(3\alpha\beta\). Rearranging:

\[\alpha^{3} + \beta^{3} = \left(\alpha + \beta\right)^{3} - 3\alpha\beta\left(\alpha + \beta\right) \qquad \textbf{M1}\]

Substituting the two known values:

\[\alpha^{3} + \beta^{3} = \left(\frac{5}{2}\right)^{3} - 3\left(2\right)\left(\frac{5}{2}\right) = \frac{125}{8} - 15 = \frac{125 - 120}{8} = \frac{5}{8} \qquad \textbf{A1}\]

An equally valid route uses the factorisation \(\alpha^{3} + \beta^{3} = \left(\alpha + \beta\right)\left(\alpha^{2} - \alpha\beta + \beta^{2}\right)\) together with part (b), giving \(\dfrac{5}{2}\left(\dfrac{9}{4} - 2\right) = \dfrac{5}{2} \times \dfrac{1}{4} = \dfrac{5}{8}\), and the agreement of the two methods is a good check. The errors to watch are cubing only the numerator, so that \(\left(\dfrac{5}{2}\right)^{3}\) becomes \(\dfrac{125}{2}\), and dropping the factor \(\left(\alpha + \beta\right)\) from the correction term, which would give \(\dfrac{125}{8} - 6\).

It is worth noting how small the final answer is compared with \(\dfrac{125}{8} \approx 15.6\). The two terms nearly cancel, which means an arithmetic slip anywhere produces a wildly wrong result rather than a slightly wrong one. Keeping everything over the denominator \(8\) rather than converting to decimals is the safest way through.

Examination takeaway. Read the sum and product of the roots straight off the coefficients, remembering the minus sign on the sum and the division by \(a\). Then express any symmetric function of the roots in terms of those two quantities using an identity written down before substitution. This works whether or not the roots are real, which is exactly why the question forbids solving the equation.