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Math 20-1 · Toolkits

Restrictions and roots that are not there

Radicals, rational expressions and absolute values all fail in the same way, for the same reason, and the fix is the same every time. Learning it once covers half the algebra in Math 20-1.

Why do equations produce answers that do not work?

Some operations lose information. Squaring loses the sign: 3 and −3 both square to 9, so once you have squared, the equation no longer knows which one it started with. Multiplying by an expression containing a variable can multiply by zero without telling you. Absolute value throws the sign away on purpose.

Every one of those makes the equation easier to solve, which is why you do it. The cost is that the easier equation is slightly bigger than the one you started with — it has a few extra solutions the original never had. They are not mistakes in your algebra. They are the algebra working correctly on a question that is no longer quite the same question.

So there is one habit that covers all of it: write down what x is allowed to be before you start, and check every answer against the original equation at the end. Nothing else in this course pays back as reliably.

Where it turns up

  • Engineering and physics — a squared equation for a time or a length routinely returns a negative root that gets discarded because time and length cannot be negative
  • Spreadsheets — a division-by-zero error is a non-permissible value appearing in real life
  • Programming — a domain check before calling a square root is exactly the restriction written down first
  • Later maths — in Math 30-1, logarithms restrict their argument to positive numbers, and the same habit applies

The words, first

The idea: Three words for closely related things, and it is worth being able to say which is which.

WordWhat it means
RestrictionA limit on the values x may take, coming from the expression itself: a radicand cannot be negative, a denominator cannot be zero.
Non-permissible valueA specific value excluded because it makes a denominator zero.
DomainThe full set of allowed inputs — what is left after the restrictions.
Extraneous rootA solution of the transformed equation that fails in the original. It is produced by the transformation, not by an error.
Equivalent equationsEquations with exactly the same solutions. Squaring does not produce an equivalent equation.

The three places it happens

The idea: Each one loses different information, and knowing which tells you what kind of false answer to expect.

SituationRestrictionWhat goes wrong
√(expression)The radicand must be ≥ 0, and the other side must be ≥ 0Squaring erases the sign, so a negative-valued solution appears
expression ÷ (denominator)The denominator ≠ 0Multiplying by the denominator can multiply by zero, so a forbidden value appears as a solution
|expression| = somethingThe right-hand side must be ≥ 0Splitting into two cases can produce a case with a negative right side, which is impossible

In all three, the transformed equation is bigger than the original. Checking is what shrinks it back.

One method, three examples

The idea: Restrictions first, solve second, check third. In that order, every time.

Radical: √(2x + 3) = x. Restrictions: 2x + 3 ≥ 0 so x ≥ −1.5, and the right side must be non-negative so x ≥ 0. Squaring gives x² − 2x − 3 = 0, so x = 3 or x = −1. The restriction x ≥ 0 rules out −1 before you even substitute. x = 3.

Rational: x/(x − 4) = 4/(x − 4) + 3. Restriction: x ≠ 4. Multiplying through gives x = 4 + 3(x − 4), so x = 4 — which is the restriction. No solution. Writing the restriction down first turns this from a surprise into an expectation.

Absolute value: |x − 3| = 2x. Restriction: the right side must be ≥ 0, so x ≥ 0. The two cases give x = −3 and x = 1, and the restriction kills the first one. x = 1.

Three different-looking questions, one procedure.

Restrictions survive simplification

The idea: Cancelling changes what the expression looks like without changing where it was undefined. The restriction has to be carried along by hand.

Consider (x² − 9)/(x + 3). Factor and cancel: (x − 3)(x + 3)/(x + 3) = x − 3.

The simplified expression is perfectly happy at x = −3 and gives −6. The original is undefined there. The two are only equal where both are defined, so the honest answer is:

x − 3,   x ≠ −3

This is worth marks on its own in almost every question that asks you to simplify, and it is the single most commonly dropped part of the answer.

Division adds restrictions. When you divide by a rational expression, the numerator of the divisor ends up in a denominator, so it cannot be zero either. That restriction exists even though it never appears in the final answer.

The habit, in one line

The idea: Before solving: what is x not allowed to be? After solving: does each answer survive the original equation?

Two questions, ten seconds each. They catch extraneous roots, non-permissible values, dropped restrictions and impossible absolute values — which between them are most of the algebra marks available in this course.

It also transfers. In Math 30-1 the same habit handles logarithms, whose argument must be positive, and rational functions, whose asymptotes are exactly the non-permissible values you learned to find here.