AS & A Level · Practice questions · Transforming polynomial equations

Change the roots, keep the algebra.

Six original Cambridge-style questions on transforming polynomial equations: reciprocal, linear and squared roots, given substitutions, and using the transformed equation to find symmetric functions quickly.

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

Write \(y\), invert to \(x\), substitute, tidy.

Choose \(y\) in terms of \(x\) from the new roots: \(y = \frac{1}{x}\), \(y = x^{2}\), \(y = x + h\), or the given one.
Invert to \(x\) in terms of \(y\). This is what you substitute.
Surds: isolate the \(\sqrt{\ }\) terms on one side, then square both whole sides.
Clear fractions and write the equation with integer coefficients.
Check the new sum of roots (or product) against the old roots.

Recall

[2 marks]

The roots of \(x^{3} + 5x^{2} - 2x + 3 = 0\) are \(\alpha\), \(\beta\), \(\gamma\). Find the cubic equation with roots \(\dfrac{1}{\alpha}\), \(\dfrac{1}{\beta}\), \(\dfrac{1}{\gamma}\).

Put \(x = \dfrac{1}{y}\): \(\dfrac{1}{y^{3}} + \dfrac{5}{y^{2}} - \dfrac{2}{y} + 3 = 0\). ✓
Multiply by \(y^{3}\): \(1 + 5y - 2y^{2} + 3y^{3} = 0\), so \(3y^{3} - 2y^{2} + 5y + 1 = 0\). ✓

Calculation

[4 marks]

The roots of \(x^{3} - 6x^{2} + 11x - 6 = 0\) are \(\alpha\), \(\beta\), \(\gamma\).

(a) Find the equation with roots \(\alpha - 2\), \(\beta - 2\), \(\gamma - 2\).

(b) Solve your new equation and hence find \(\alpha\), \(\beta\) and \(\gamma\).

(a) Put \(x = y + 2\): \((y + 2)^{3} - 6(y + 2)^{2} + 11(y + 2) - 6 = 0\). ✓
\(y^{3} + 6y^{2} + 12y + 8 - 6y^{2} - 24y - 24 + 11y + 22 - 6 = 0\), so \(y^{3} - y = 0\). ✓
(b) \(y(y - 1)(y + 1) = 0\), so \(y = -1, 0, 1\). ✓
Add 2: \(\alpha, \beta, \gamma = 1, 2, 3\). ✓

Calculation

[5 marks]

The roots of \(x^{3} - 2x^{2} + 4x + 1 = 0\) are \(\alpha\), \(\beta\), \(\gamma\).

(a) Show that the equation with roots \(\alpha^{2}\), \(\beta^{2}\), \(\gamma^{2}\) is \(y^{3} + 4y^{2} + 20y - 1 = 0\).

(b) Hence find \(\alpha^{4} + \beta^{4} + \gamma^{4}\).

(a) \(x = \sqrt{y}\): \(y\sqrt{y} - 2y + 4\sqrt{y} + 1 = 0\), so \(\sqrt{y}\,(y + 4) = 2y - 1\). ✓
Square: \(y(y + 4)^{2} = (2y - 1)^{2}\), so \(y^{3} + 8y^{2} + 16y = 4y^{2} - 4y + 1\). ✓
Hence \(y^{3} + 4y^{2} + 20y - 1 = 0\). ✓
(b) For the new roots \(p = \alpha^{2}\) etc.: \(\sum p = -4\), \(\sum pq = 20\). ✓
\(\sum \alpha^{4} = \sum p^{2} = (\sum p)^{2} - 2\sum pq = 16 - 40 = -24\). ✓

Analysis

[5 marks]

The roots of \(x^{3} - 3x^{2} + x + 2 = 0\) are \(\alpha\), \(\beta\), \(\gamma\).

(a) Use the substitution \(y = \dfrac{x}{x + 1}\) to show that \(\dfrac{\alpha}{\alpha + 1}\), \(\dfrac{\beta}{\beta + 1}\), \(\dfrac{\gamma}{\gamma + 1}\) are the roots of \(3y^{3} + y^{2} - 5y + 2 = 0\).

(b) Hence find \(\dfrac{\alpha + 1}{\alpha} + \dfrac{\beta + 1}{\beta} + \dfrac{\gamma + 1}{\gamma}\).

(a) \(yx + y = x\), so \(x(1 - y) = y\) and \(x = \dfrac{y}{1 - y}\). ✓
\(\dfrac{y^{3}}{(1 - y)^{3}} - \dfrac{3y^{2}}{(1 - y)^{2}} + \dfrac{y}{1 - y} + 2 = 0\); multiply by \((1 - y)^{3}\). ✓
\(y^{3} - 3y^{2}(1 - y) + y(1 - y)^{2} + 2(1 - y)^{3} = y^{3} - 3y^{2} + 3y^{3} + y - 2y^{2} + y^{3} + 2 - 6y + 6y^{2} - 2y^{3}\). ✓
\(= 3y^{3} + y^{2} - 5y + 2\), so the equation is \(3y^{3} + y^{2} - 5y + 2 = 0\). ✓
(b) The required sum is \(\sum \dfrac{1}{y}\) over the new roots \(= \dfrac{\sum pq}{pqr} = \dfrac{-5/3}{-2/3} = \dfrac{5}{2}\). ✓
Check directly: \(\sum\left(1 + \dfrac{1}{\alpha}\right) = 3 + \dfrac{\sum \alpha\beta}{\alpha\beta\gamma} = 3 + \dfrac{1}{-2} = \dfrac{5}{2}\). ✓

Analysis

[5 marks]

The roots of \(x^{3} + 2x^{2} - 1 = 0\) are \(\alpha\), \(\beta\), \(\gamma\).

(a) Using the substitution \(y = \dfrac{1}{x^{2}}\), show that \(\dfrac{1}{\alpha^{2}}\), \(\dfrac{1}{\beta^{2}}\), \(\dfrac{1}{\gamma^{2}}\) are the roots of \(y^{3} - 4y^{2} + 4y - 1 = 0\).

(b) Hence find \(\dfrac{1}{\alpha^{4}} + \dfrac{1}{\beta^{4}} + \dfrac{1}{\gamma^{4}}\).

(a) \(x = \dfrac{1}{\sqrt{y}}\): \(\dfrac{1}{y\sqrt{y}} + \dfrac{2}{y} - 1 = 0\). Multiply by \(y\sqrt{y}\): \(1 + 2\sqrt{y} - y\sqrt{y} = 0\). ✓
Isolate: \(\sqrt{y}\,(2 - y) = -1\). Square: \(y(2 - y)^{2} = 1\). ✓
\(y^{3} - 4y^{2} + 4y - 1 = 0\). ✓
(b) New roots \(p, q, r\): \(\sum p = 4\), \(\sum pq = 4\). ✓
\(\sum \dfrac{1}{\alpha^{4}} = \sum p^{2} = 4^{2} - 2(4) = 8\). ✓

Proof

[6 marks]

The roots of \(x^{3} + 3x - 1 = 0\) are \(\alpha\), \(\beta\), \(\gamma\).

(a) Use the substitution \(y = x^{2} + 1\) to find a cubic equation with roots \(\alpha^{2} + 1\), \(\beta^{2} + 1\), \(\gamma^{2} + 1\).

(b) Find the value of \(\dfrac{1}{\alpha^{2} + 1} + \dfrac{1}{\beta^{2} + 1} + \dfrac{1}{\gamma^{2} + 1}\).

(a) \(x = \sqrt{y - 1}\): \((y - 1)\sqrt{y - 1} + 3\sqrt{y - 1} - 1 = 0\), so \(\sqrt{y - 1}\,(y + 2) = 1\). ✓
Square: \((y - 1)(y + 2)^{2} = 1\), so \(y^{3} + 3y^{2} - 4 = 1\), giving \(y^{3} + 3y^{2} - 5 = 0\). ✓
(b) The sum is \(\dfrac{\sum pq}{pqr}\) for the new roots: \(\sum pq = 0\) and \(pqr = 5\), so the sum is \(0\). ✓
(c) If all three roots were real, each \(\alpha^{2} + 1 \ge 1 > 0\), so each term \(\dfrac{1}{\alpha^{2} + 1}\) would be positive. ✓
Three positive terms cannot add to \(0\), so at least one root is not real (in fact two, as complex roots of a real cubic come in conjugate pairs). ✓

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.