Solving Equations and Checking Solutions, foundations for calculus lesson

In calculus you find the highest or lowest point of a graph by solving an equation, and a single false root can send you to an input where nothing special happens, or to one outside the domain. Solving well means knowing, at each line, whether the new equation has exactly the same solutions as the old one. When a step cannot promise that, the fix is simple. Treat the results as candidates and test each one in the equation you started with.

Key terms

Equation and solution

Definition An equation in the real variable $x$ is a statement $A(x) = B(x)$ between two expressions. A solution is a real number $c$ at which both $A(c)$ and $B(c)$ are defined and $A(c) = B(c)$ is true. The solution set is the set of all solutions, which may be empty.

In plain words A solution is an input that makes both sides exist and come out equal.

Example The equation $2x + 3 = 11$ has solution set $\{4\}$ , since $2 \cdot 4 + 3 = 11$ . The equation $x^2 = x$ has solution set $\{0, 1\}$ , and $x + 1 = x$ has no solution at all.

Equivalent step

Definition Two equations are equivalent when they have the same solution set. The following steps produce an equivalent equation.

Add or subtract the same real quantity on both sides, when that quantity is defined at every input where the equation is.
Multiply or divide both sides by the same nonzero constant.

Multiplying or dividing both sides by a variable expression $C(x)$ is reversible only at inputs where $C(x)$ is defined and $C(x) \ne 0$ , so it needs that condition stated separately. For a product, the zero product property says that for real numbers $AB = 0$ exactly when $A = 0$ or $B = 0$ , so at least one factor is zero. Use it only once one side of the equation is $0$ and the other side is written as a product.

In plain words An equivalent step can be undone, so it neither adds nor loses solutions. Dividing by something that might be zero can throw solutions away.

Example Dividing $x^2 = 3x$ by $x$ gives $x = 3$ and silently loses $x = 0$ . Instead, move everything to one side.

\begin{aligned} &x^2 - 3x = 0 \\[0.4em] \Longleftrightarrow \quad &x(x - 3) = 0 \\[0.4em] \Longleftrightarrow \quad &x = 0 \ \text{ or } \ x = 3. \end{aligned}

The zero product property needs a zero on one side. From $(x - 1)(x - 2) = 6$ you may not conclude $x - 1 = 6$ . Expanding gives $x^2 - 3x - 4 = 0$ , whose solutions are $-1$ and $4$ .

Restriction

Definition The restrictions of an equation are the conditions under which both sides are defined, such as every denominator being nonzero and every real even root having a nonnegative radicand. Only numbers that satisfy all restrictions can be solutions, so the restrictions are found from the original equation before any step is taken.

In plain words Write down the forbidden inputs first. Anything later that lands on one of them is thrown out.

Example In the equation

\frac{x}{x + 2} - \frac{1}{x - 2} = \frac{-4}{x^2 - 4}

the denominators require $x \ne 2$ and $x \ne -2$ . These two numbers cannot be solutions, whatever the algebra later produces.

Candidate solution

Definition A candidate solution is a number produced by a chain of steps in which at least one step is not known to be reversible, for example multiplying by a variable expression or squaring. Say that a step keeps all solutions when every solution of the equation before the step is also a solution of the equation after it. Equivalent steps keep all solutions. So do squaring both sides and multiplying both sides by an expression that is defined at every input allowed by the restrictions, even where that expression is zero. Dividing both sides by an expression, or cancelling a factor, does not, because a solution at which that expression is zero can be lost. If every step in the chain keeps all solutions, then every solution of the original equation is among the candidates. A candidate still need not be a solution, which is why each one is checked.

In plain words Candidates are suspects. Some are real solutions and some are side effects of the method.

Example Return to the equation above and multiply both sides by $(x - 2)(x + 2)$ , which may be zero.

\begin{aligned} x(x - 2) - (x + 2) &= -4 \\[0.4em] x^2 - 3x + 2 &= 0 \\[0.4em] (x - 1)(x - 2) &= 0. \end{aligned}

The candidates are $x = 1$ and $x = 2$ . The restriction rules out $2$ at once. For $x = 1$ the left side is $\frac{1}{3} + 1 = \frac{4}{3}$ and the right side is $\frac{-4}{-3} = \frac{4}{3}$ , so the solution set is $\{1\}$ .

Squaring

Definition If $A = B$ , then $A^2 = B^2$ . The converse is false, because $A^2 = B^2$ holds exactly when $A = B$ or $A = -B$ . So squaring both sides gives a necessary condition for a solution, and it may introduce extra candidates that solve $A = -B$ instead.

In plain words Squaring hides the sign. A number that made the two sides opposite will look like a solution after squaring.

Example The equation $x = 3$ has one solution, but the squared equation $x^2 = 9$ also accepts $x = -3$ , which solves $x = -3$ rather than $x = 3$ .

Checking in the original equation

Definition To check a candidate $c$ , substitute it into the original equation and confirm that both sides are defined at $c$ and equal. Substituting into a later line is not a check, because that line may already contain the extra candidates. After squaring or after multiplying by a variable expression this check is a required part of the solution, not an optional extra.

In plain words Go back to the very first line and test each suspect there.

Example Solve $\sqrt{x + 7} = x - 5$ . The radicand requires $x \ge -7$ . Squaring gives

\begin{aligned} x + 7 &= x^2 - 10x + 25, \\[0.4em] \text{so} \ \ x^2 - 11x + 18 &= (x - 2)(x - 9) \\[0.4em] &= 0. \end{aligned}

The candidates are $2$ and $9$ . Check both in the original.

Candidate	Left side $\sqrt{x + 7}$	Right side $x - 5$	Solution?
$x = 9$	$4$	$4$	yes
$x = 2$	$3$	$-3$	no

The candidate $2$ solves $\sqrt{x + 7} = -(x - 5)$ , which the squaring step could not tell apart. The solution set is $\{9\}$ .

Check yourself

Solve $\sqrt{2x + 3} = x$ and check every candidate.

Show the answer

Squaring gives $2x + 3 = x^2$ , so $x^2 - 2x - 3 = (x - 3)(x + 1) = 0$ and the candidates are $3$ and $-1$ . At $x = 3$ both sides equal $3$ . At $x = -1$ the left side is $\sqrt{1} = 1$ but the right side is $-1$ . The solution set is $\{3\}$ .

Solve the equation below.

\frac{x}{x - 1} = \frac{1}{x - 1} + 2

Show the answer

The restriction is $x \ne 1$ . Multiplying by $x - 1$ gives $x = 1 + 2(x - 1)$ , so $x = 1$ . That candidate is excluded, so the equation has no solution.

A student divides $x^3 = 4x$ by $x$ and reports $x = \pm 2$ . What is the full solution set?

Show the answer

Write $x^3 - 4x = x(x - 2)(x + 2) = 0$ . The solution set is $\{-2, 0, 2\}$ . Dividing by $x$ lost $x = 0$ , since that step assumed $x \ne 0$ .

Solving Equations and Checking Solutions

Key terms

Equation and solution

Equivalent step

Restriction

Candidate solution

Squaring

Checking in the original equation

Check yourself

How this lesson connects

Builds on

Used later in

Where this leads in Calculus