Chapter 1 — Fundamental Concepts¶
1.5 — Equations¶
To solve an equation is to ask for which values a statement stops being a possibility and becomes a fact.
An equation is an equality between two expressions in which at least one variable appears. Unlike an identity, such as those of §1.3, it is in general not true for every value of the variable, but only for some, or for none. Each value that makes it true is a solution, and the set of all of them is the solution set. To solve the equation is to find that set in full: it is not enough to find some solutions; we must make sure that none has been left out and that nothing has slipped in that is not one.
With the factoring of §1.3 and the rational expressions of §1.4, this section organizes the methods of solution according to the type of expression. The thread that ties them together is a single question: when does an algebraic step preserve the solution set, and when does it not? From the answer comes a phenomenon that appears here for the first time, the extraneous solution: a number that solves an equation obtained along the way, but not the original one.
§1. Equivalent equations and linear equations¶
An equation in \(x\) has the form \(A(x) = B(x)\), where \(A\) and \(B\) are algebraic expressions. It only makes sense to ask whether a number \(r\) is a solution when both sides are defined at \(r\); the set of those numbers is the domain of the equation, in the same sense as in §1.4. Two equations are equivalent when they have the same solution set. Solving an equation consists, almost always, in replacing it by an equivalent and simpler one, until we reach one whose solution set is plain to see. The two basic steps that guarantee equivalence are the following.
Proposition. Let \(A(x) = B(x)\) be an equation with domain \(D\), and let \(C(x)\) be an expression defined on all of \(D\). Then, for \(r \in D\):
(i) \(r\) is a solution of \(A(x) = B(x)\) if and only if it is a solution of \(A(x) + C(x) = B(x) + C(x)\);
(ii) if moreover \(C(r) \neq 0\), then \(r\) is a solution of \(A(x) = B(x)\) if and only if it is a solution of \(A(x)C(x) = B(x)C(x)\).
Proof. Fix \(r \in D\) and write \(a = A(r)\), \(b = B(r)\) and \(c = C(r)\), which are real numbers. For (i): if \(a = b\), adding the same number \(c\) to both sides gives \(a + c = b + c\); conversely, if \(a + c = b + c\), adding \(-c\) to both sides gives \(a = b\). For (ii): if \(a = b\), then \(ac = bc\); conversely, if \(ac = bc\) and \(c \neq 0\), multiplying by the inverse \(c^{-1}\) of §1.1 gives \(a = b\). \(\blacksquare\)
The key lies in the "conversely". Each of these steps can be undone by another step of the same kind, and that is why it neither gains nor loses solutions. Multiplying by an expression that vanishes at some point, on the other hand, can make new solutions appear: the passage from \(ac = bc\) to \(a = b\) fails when \(c = 0\). This is the origin of all the extraneous solutions in this section.
A linear equation in \(x\) is one that, by means of those steps, can be brought to the form \(ax + b = 0\), with \(a, b \in \mathbb{R}\). If \(a \neq 0\), we add \(-b\) and multiply by \(a^{-1}\), and we obtain the equivalent equation \(x = -\frac{b}{a}\): there is a unique solution. If \(a = 0\), the equation reduces to \(b = 0\), a statement in which \(x\) no longer appears. It is true for every \(x\) if \(b = 0\), and then the solution set is \(\mathbb{R}\), or false for every \(x\) if \(b \neq 0\), and then the solution set is empty, \(\emptyset\). This case is not an oddity: it reminds us that solving does not always produce a single number.
Reminder. Adding the same expression to both sides, or multiplying them by an expression that does not vanish, yields an equivalent equation. The equation \(ax + b = 0\) has a unique solution if \(a \neq 0\); if \(a = 0\), the solution set is \(\mathbb{R}\) or \(\emptyset\), according as \(b = 0\) or \(b \neq 0\).
§2. Quadratic equations¶
A quadratic equation has the form \(ax^2 + bx + c = 0\), with \(a \neq 0\). The heavy lifting was already done in §1.3, which factored the trinomial by completing the square. Only one ingredient is missing here, the one that lets us pass from a zero product to its factors.
Zero product property. For \(u, v \in \mathbb{R}\), \(uv = 0\) if and only if \(u = 0\) or \(v = 0\).
One of the two implications is the lemma of §1.3: if \(uv = 0\), one of the factors is zero. The other comes from §1.1: if \(u = 0\), then \(uv = 0 \cdot v = 0\), and likewise if \(v = 0\). This property, and not a habit of manipulation, is what allows us to pass from \((x - 3)(x + 2) = 0\) to "\(x = 3\) or \(x = -2\)". The proof of the lemma uses the fact that every nonzero real has an inverse; in a number system without that guarantee, the property could fail. With a product of three or more factors it is applied several times: the product is zero if and only if one of the factors is.
Proposition. Let \(a, b, c \in \mathbb{R}\) with \(a \neq 0\), and let \(\Delta = b^2 - 4ac\) be the discriminant.
(i) If \(\Delta > 0\), the equation \(ax^2 + bx + c = 0\) has exactly two solutions,
(ii) If \(\Delta = 0\), it has exactly one solution, \(x = -\dfrac{b}{2a}\).
(iii) If \(\Delta < 0\), it has no real solutions.
Proof. If \(\Delta \ge 0\), §1.3 gives \(ax^2 + bx + c = a(x - x_1)(x - x_2)\) for every \(x\). By the zero product property, and since \(a \neq 0\), the right-hand side equals \(0\) if and only if \(x = x_1\) or \(x = x_2\). Moreover \(x_1 - x_2 = \frac{\sqrt{\Delta}}{a}\), which is nonzero if \(\Delta > 0\) and zero if \(\Delta = 0\). This proves (i) and (ii), because if \(\Delta = 0\) both formulas give \(-\frac{b}{2a}\). If \(\Delta < 0\), in the proof in §1.3 we saw that \(at^2 + bt + c \neq 0\) for every real \(t\), and that is (iii). \(\blacksquare\)
The formula that sums up (i) and (ii) is written with a single symbol, \(x = \dfrac{-b \pm \sqrt{\Delta}}{2a}\), where \(\pm\) indicates that we take the sign \(+\) once and the sign \(-\) once. For example, in \(x^2 - 4x + 1 = 0\) we have \(\Delta = 16 - 4 = 12\) and \(\sqrt{12} = 2\sqrt{3}\), so that \(x = \frac{4 \pm 2\sqrt{3}}{2} = 2 \pm \sqrt{3}\). Factoring and using the formula are not two different methods but the same result read in two ways: when the factorization is visible at a glance, it is faster; when it is not, the formula always finds it.
Reminder. A product is zero if and only if one of its factors is. The sign of \(\Delta\) decides whether the quadratic equation has two solutions, one, or none.
§3. Equations with radicals¶
An equation with radicals is one in which the variable appears inside a root, as in \(\sqrt{x + 3} = x - 3\). The usual method is to isolate the radical and raise both sides to the index of the root to eliminate it. But squaring is not one of the safe steps of §1: it cannot be undone, because the square does not distinguish a number from its opposite. Indeed, by the difference of squares of §1.3 and the zero product property,
The symbol \(\iff\) reads "if and only if": each statement implies the next one and also the previous one. Squaring the equality \(u = v\) produces a weaker statement, which also holds when \(u = -v\). Extraneous solutions come from there, and the following proposition says exactly which ones they are.
Proposition. Let \(f(x)\) and \(g(x)\) be algebraic expressions. A real number \(r\) is a solution of \(\sqrt{f(x)} = g(x)\) if and only if \(f(r) = g(r)^2\) and \(g(r) \ge 0\).
Proof. Suppose that \(\sqrt{f(r)} = g(r)\). A square root is, by definition (§1.2), a nonnegative number, so \(g(r) \ge 0\); and squaring, \(f(r) = \left(\sqrt{f(r)}\right)^2 = g(r)^2\). Conversely, suppose that \(f(r) = g(r)^2\) and \(g(r) \ge 0\). Then \(f(r) \ge 0\), because it is a square, and \(\sqrt{f(r)}\) is defined. Moreover, \(g(r)\) is a nonnegative number whose square is \(f(r)\), and by the uniqueness of the root proved in §1.2, \(\sqrt{f(r)} = g(r)\). \(\blacksquare\)
The proposition says that every solution of the original equation solves the squared equation, \(f(x) = g(x)^2\), but that the latter may have extra solutions: those with \(g(r) < 0\). Those are the extraneous solutions. They are not computational errors, but the necessary consequence of a step that loses information. The practical conclusion is that every solution obtained by squaring must be checked in the original equation, or else by verifying that \(g(r) \ge 0\). The same happens with any even index. When there is more than one radical and we must square twice, the safest course is to check each candidate in the starting equation.
Example. Let us solve \(\sqrt{x + 3} = x - 3\). Squaring, \(x + 3 = (x - 3)^2 = x^2 - 6x + 9\), that is, \(x^2 - 7x + 6 = 0\), which factors as \((x - 1)(x - 6) = 0\). The candidates are \(x = 1\) and \(x = 6\). Here \(g(x) = x - 3\). For \(x = 6\), \(g(6) = 3 \ge 0\), and indeed \(\sqrt{9} = 3\). For \(x = 1\), \(g(1) = -2 < 0\): it is an extraneous solution, because \(\sqrt{4} = 2\) and not \(-2\). The solution set is \(\{6\}\).
Reminder. Squaring can add solutions, because \(u^2 = v^2\) also holds when \(u = -v\). Every solution of an equation with radicals is checked in the original equation.
§4. Equations with rational expressions¶
When the variable appears in a denominator, the usual procedure is to multiply both sides by the least common denominator of §1.4. The fractions disappear and the equation reduces to a linear or quadratic one. Is this a safe step? By part (ii) of the proposition of §1, it is on the whole domain of the original equation, because there the least common denominator does not vanish. Outside that domain the proposition says nothing. The equation without fractions may have solutions at the excluded values, and those are not solutions of the original equation, which is not even defined there.
From this comes the method. First we determine the domain of the original equation. Then we multiply by the least common denominator and solve the resulting equation. Finally, we keep only the solutions that belong to the domain. It is the same care that simplification required in §1.4: the expression without denominators does not remember which values were forbidden.
Example. Let us solve
Since \(x^2 - 1 = (x - 1)(x + 1)\), the domain excludes \(x = 1\) and \(x = -1\), and the least common denominator is \((x - 1)(x + 1)\). Multiplying both sides by it leaves \((x + 1) + (x - 1) = 2\), that is, \(2x = 2\), whose only solution is \(x = 1\). But \(1\) is not in the domain. The original equation has no solutions: its solution set is \(\emptyset\).
The opposite phenomenon also occurs. In \(\dfrac{x}{x - 2} = \dfrac{2}{x - 2} + 1\), the domain excludes \(x = 2\). Multiplying by \(x - 2\) we obtain \(x = 2 + (x - 2)\), that is, \(x = x\), which holds for every real. Discarding the excluded value, the solution set is \(\mathbb{R} \setminus \{2\}\), and not all of \(\mathbb{R}\) as the equation without denominators would suggest.
Reminder. Multiplying by the least common denominator yields an equivalent equation only within the domain of the original. Every solution is checked against that domain, and discarded if it falls outside.
With the methods of solution now established, the chapter is ready for their most direct application: translating problems described in words into the language of equations, the task of the next section.