Theory of Equations: One Mark Questions, Book Back Answers and Notes

About Theory of Equations

Medium ~120 min study

Many practical questions reduce to a polynomial equation. If an open box is made from a square sheet by cutting squares from the corners, asking for a given volume leads to a cubic equation. This chapter gives you the tools to handle such equations without always solving them completely.

The central idea is the link between the roots of a polynomial and its coefficients. For a quadratic you already know that the sum of the roots is $-\frac{b}{a}$ and the product is $\frac{c}{a}$. Vieta's formulae extend this to any degree. The Fundamental Theorem of Algebra tells you how many roots to look for, and your work on complex numbers from Unit 2 explains why non-real roots come in pairs.

In the exam you will be asked to form an equation from given roots, find the remaining roots when one root such as $2 + i$ or $1 + \sqrt{3}$ is given, solve reciprocal equations, and use Descartes' rule of signs to say how many positive and negative roots are possible. The skill that helps most is writing down Vieta's relations carefully, with the right signs, before you start.

What you'll learn

• Write down the sum and products of roots for cubic and higher degree equations.
• Form a polynomial equation from given roots, or from roots related to those of another equation.
• Find the missing roots when one non-real complex root or one surd root is given.
• Recognise reciprocal equations and solve them by the substitution $y = x + \frac{1}{x}$.
• List possible rational roots with the rational root theorem and test them.
• Use Descartes' rule of signs to find the possible numbers of positive, negative and non-real roots.

Before you start

• Solving quadratic equations and the relations between their roots and coefficients.
• Complex numbers and conjugates from Unit 2.
• The factor theorem, the remainder theorem and synthetic division.

Theory of Equations explained

Polynomial Equations and Their Roots

Basics of Polynomial Equations

A polynomial of degree $n$ is $P(x) = a_n x^n + a_{n-1}x^{n-1} + \cdots + a_1 x + a_0$ with $a_n \neq 0$. A value $c$ with $P(c) = 0$ is a zero of the polynomial and a root of the equation $P(x) = 0$. If $a_n = 1$ the polynomial is monic. A root $c$ has multiplicity $m$ when $(x - c)^m$ divides $P(x)$ but $(x - c)^{m+1}$ does not.

Vieta's Formulae and Forming Equations

For a cubic $ax^3 + bx^2 + cx + d = 0$ with roots $\alpha, \beta, \gamma$: $\sum \alpha = -\frac{b}{a}$, $\sum \alpha\beta = \frac{c}{a}$ and $\alpha\beta\gamma = -\frac{d}{a}$. The signs alternate, starting with minus. The same pattern holds for any degree. Going the other way, the monic cubic with roots $\alpha, \beta, \gamma$ is $x^3 - (\sum \alpha)x^2 + (\sum \alpha\beta)x - \alpha\beta\gamma = 0$. This also lets you build an equation whose roots are, say, the squares or reciprocals of the roots of a given equation, without solving the original.

The Fundamental Theorem and the Nature of Roots

The Fundamental Theorem of Algebra says every polynomial equation of degree $n \geq 1$ has at least one complex root. It follows that it has exactly $n$ complex roots, counted with multiplicity. Two pairing results cut the work down. If the coefficients are real and $a + ib$ ($b \neq 0$) is a root, then $a - ib$ is also a root. If the coefficients are rational and $p + \sqrt{q}$ is a root, with $p, q$ rational and $\sqrt{q}$ irrational, then $p - \sqrt{q}$ is also a root. Given one such root, you get a quadratic factor straight away.

Solving Higher Degree Equations

The rational root theorem says: if $\frac{p}{q}$, in lowest terms, is a root of a polynomial with integer coefficients, then $p$ divides $a_0$ and $q$ divides $a_n$. This gives a short list of candidates to test. A polynomial is reciprocal when $x^n P\left(\frac{1}{x}\right) = \pm P(x)$; then whenever $\alpha$ is a root, so is $\frac{1}{\alpha}$. For an even-degree reciprocal equation with $a_k = a_{n-k}$, divide by $x^{n/2}$ and substitute $y = x + \frac{1}{x}$ to halve the degree. Some non-polynomial equations, such as those involving $\sqrt{x}$ or $e^x$, can be changed into polynomial equations by a suitable substitution. Check the answers in the original equation, since some may not be valid.

Descartes' Rule of Signs

Write $P(x)$ in descending powers and count the sign changes between consecutive non-zero coefficients. The number of positive roots is at most this count. Do the same with $P(-x)$ to get the maximum number of negative roots. For example, $P(x) = x^3 - 2x^2 + x - 5$ has three sign changes, so at most three positive roots. $P(-x) = -x^3 - 2x^2 - x - 5$ has none, so there are no negative roots. Zero is not a root either, so the three roots are either all positive or one positive root with a pair of non-real roots.

Common mistakes to avoid

• Counting sign changes in Descartes' rule before arranging the terms in descending powers of $x$.
• Pairing $p + \sqrt{q}$ with $p - \sqrt{q}$ when the coefficients are not rational. The pairing rule needs rational coefficients, just as the conjugate rule needs real ones.
• Losing count of repeated roots. If $(x - 2)^2$ is a factor, the root 2 counts twice towards the total of $n$ roots.
• Getting the sign wrong in Vieta's formulae. For $ax^3 + bx^2 + cx + d = 0$ the sum of roots is $-\frac{b}{a}$, not $\frac{b}{a}$, and the product is $-\frac{d}{a}$.

Test yourself on these with the practice test, then check the worked reasoning in the solved MCQs.

Frequently asked questions

What is the difference between a zero and a root?

A zero belongs to a polynomial: it is a value $c$ with $P(c) = 0$. A root belongs to an equation: it is a solution of $P(x) = 0$. They are the same numbers, and in practice the two words are used interchangeably.

Can a cubic equation have only one real root?

Yes. If the coefficients are real, non-real roots come in conjugate pairs, so a cubic has either three real roots or one real root and a pair of complex conjugate roots. Either way, a cubic with real coefficients always has at least one real root. For example, $x^3 - 1 = 0$ has the real root 1 and the pair $\omega, \omega^2$. With non-real coefficients this guarantee does not hold.

How do I know if an equation is reciprocal?

Look at the coefficients from both ends. If they match ($a_k = a_{n-k}$) or match with opposite signs ($a_k = -a_{n-k}$), the equation is reciprocal. In both cases, if $\alpha$ is a root then so is $\frac{1}{\alpha}$. An odd-degree reciprocal equation of the first type has $-1$ as a root, and one of the second type has $1$ as a root.

Does Descartes' rule give the exact number of roots?

No. It gives only the maximum number of positive roots and the maximum number of negative roots. The actual number of positive roots is either the number of sign changes or less than it by an even number, and the same holds for negative roots.

When can I use the rational root theorem?

When all the coefficients are integers. List the factors of the constant term and of the leading coefficient, form every fraction $\frac{p}{q}$ with both signs, and test these values. For $2x^3 - x^2 - 2x + 1 = 0$, the candidates are $\pm 1$ and $\pm \frac{1}{2}$.

What does it mean for a root to have multiplicity?

It tells you how many times the factor for that root appears. In $(x - 1)^3(x + 2) = 0$, the root 1 has multiplicity 3 and $-2$ has multiplicity 1, so the equation of degree 4 has four roots counted with multiplicity.