AS & A Level · Practice questions · Sums and products of roots

Coefficients that know the roots.

Six original Cambridge-style questions on the relations between roots and coefficients: writing down the sums and products, forming equations, and finding unknown coefficients from what you are told about the roots.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Name the roots, write the relations, solve.

Read \(a, b, c, d\) (and \(e\)) with their signs; a missing term has coefficient 0.
Relations: \(\sum \alpha = -\frac{b}{a}\), \(\sum \alpha\beta = \frac{c}{a}\), \(\sum \alpha\beta\gamma = -\frac{d}{a}\), and the product of all the roots is \(\frac{e}{a}\) for a quartic (\(-\frac{d}{a}\) for a cubic).
Name the roots to fit the wording: \(a - d, a, a + d\); \(\alpha, 2\alpha, \beta\); a conjugate pair \(p \pm qi\).
Build an equation with alternating signs, then clear fractions.
Check by substituting a root you found, or by expanding the factors.

Recall

[3 marks]

The roots of \(4x^{3} - 8x^{2} + 5x + 3 = 0\) are \(\alpha, \beta, \gamma\). Write down the values of \(\alpha + \beta + \gamma\), \(\alpha\beta + \beta\gamma + \gamma\alpha\) and \(\alpha\beta\gamma\).

\(\sum \alpha = -\dfrac{-8}{4} = 2\). ✓
\(\sum \alpha\beta = \dfrac{5}{4}\). ✓
\(\alpha\beta\gamma = -\dfrac{3}{4}\). ✓

Calculation

[3 marks]

The roots of \(2x^{2} - 6x + 1 = 0\) are \(\alpha\) and \(\beta\). Without solving the equation, find the values of (a) \(\alpha^{2} + \beta^{2}\) (b) \((\alpha - \beta)^{2}\).

\(\alpha + \beta = 3\) and \(\alpha\beta = \dfrac{1}{2}\). ✓
(a) \(\alpha^{2} + \beta^{2} = (\alpha + \beta)^{2} - 2\alpha\beta = 9 - 1 = 8\). ✓
(b) \((\alpha - \beta)^{2} = \alpha^{2} + \beta^{2} - 2\alpha\beta = 8 - 1 = 7\). ✓

Calculation

[3 marks]

Find a cubic equation with integer coefficients whose roots are \(\dfrac{1}{2}\), \(-3\) and \(2\).

\(\sum \alpha = \dfrac{1}{2} - 3 + 2 = -\dfrac{1}{2}\). ✓
\(\sum \alpha\beta = -\dfrac{3}{2} + 1 - 6 = -\dfrac{13}{2}\); \(\alpha\beta\gamma = -3\). ✓
\(x^{3} + \dfrac{1}{2}x^{2} - \dfrac{13}{2}x + 3 = 0\); multiply by 2: \(2x^{3} + x^{2} - 13x + 6 = 0\). ✓

Analysis

[5 marks]

The roots of \(8x^{3} - 36x^{2} + kx - 15 = 0\) are in arithmetic progression.

(a) Show that one root is \(\dfrac{3}{2}\).

(b) Find the value of \(k\) and the other two roots.

(a) Let the roots be \(a - d,\ a,\ a + d\). Then \(3a = \dfrac{36}{8} = \dfrac{9}{2}\), so \(a = \dfrac{3}{2}\). ✓
(b) Substitute \(x = \dfrac{3}{2}\): \(8 \cdot \dfrac{27}{8} - 36 \cdot \dfrac{9}{4} + \dfrac{3k}{2} - 15 = 27 - 81 + \dfrac{3k}{2} - 15 = 0\). ✓
\(\dfrac{3k}{2} = 69\), so \(k = 46\). ✓
Product: \(\left(\dfrac{9}{4} - d^{2}\right)\dfrac{3}{2} = \dfrac{15}{8}\), so \(\dfrac{9}{4} - d^{2} = \dfrac{5}{4}\) and \(d = \pm 1\). ✓
The other roots are \(\dfrac{1}{2}\) and \(\dfrac{5}{2}\). ✓

Analysis

[6 marks]

The quartic equation \(x^{4} - 2x^{3} + px^{2} + qx + 6 = 0\), where \(p\) and \(q\) are real, has a root \(1 + i\).

(a) Write down a second root.

(b) The other two roots are \(\gamma\) and \(\delta\). Find \(\gamma + \delta\) and \(\gamma\delta\), and hence find \(\gamma\) and \(\delta\).

(a) The coefficients are real, so \(1 - i\) is also a root. This pair has sum 2 and product 2. ✓
(b) \(\sum \alpha = 2\): \(2 + \gamma + \delta = 2\), so \(\gamma + \delta = 0\). Product: \(2\gamma\delta = 6\), so \(\gamma\delta = 3\). ✓
\(\gamma\) and \(\delta\) are the roots of \(x^{2} + 3 = 0\), so they are \(\pm i\sqrt{3}\). ✓
(c) \(p = \sum \alpha\beta = 2 + \gamma\delta + (2)(\gamma + \delta) = 2 + 3 + 0 = 5\). ✓
\(\sum \alpha\beta\gamma = 2(\gamma + \delta) + \gamma\delta(2) = 0 + 6 = 6 = -q\), so \(q = -6\). ✓
Check: \((x^{2} - 2x + 2)(x^{2} + 3) = x^{4} - 2x^{3} + 5x^{2} - 6x + 6\). ✓

Analysis

[5 marks]

The equation \(x^{3} - 7x^{2} + k = 0\), where \(k \ne 0\), has one root that is twice another. Find the roots and the value of \(k\).

Let the roots be \(\alpha,\ 2\alpha,\ \beta\). Sum: \(3\alpha + \beta = 7\). ✓
There is no \(x\) term, so \(\sum \alpha\beta = 2\alpha^{2} + 3\alpha\beta = 0\), giving \(\alpha(2\alpha + 3\beta) = 0\). ✓
\(\alpha = 0\) would make 0 a root and \(k = 0\), so \(2\alpha = -3\beta\). With \(\beta = 7 - 3\alpha\): \(2\alpha = -21 + 9\alpha\), so \(\alpha = 3\) and \(\beta = -2\). ✓
The roots are \(3, 6, -2\). ✓
\(\alpha\beta\gamma = (3)(6)(-2) = -36 = -k\), so \(k = 36\). ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.