Transforming an equation often produces candidates, not a finished answer. A number may make an original denominator zero or have the wrong sign after squaring. Checking returns us to the question originally asked.
We use rational expressions, roots, systems and quadratic equations. We work over the real numbers.
An unknown time in the denominator
One pump fills a tank in 6 hours. Together with a second, they take 2 hours. Assume constant independent flows, no leaks and simultaneous starts. Let x > 0 be the second pump’s time alone in hours. Their rates give:
(1)/(6) + (1)/(x) = (1)/(2).
This is a rational equation with the unknown in a denominator. The original domain requires x ≠ 0; the physical condition requires x > 0. Multiply by 6x, nonzero on this domain: x + 6 = 3x, hence x = 3 hours. Checking 1/6 + 1/3 = 1/2 confirms the rate, and positive time confirms the answer’s meaning.
Cancellation can hide an empty solution set
Consider ((x)^(2) − 1)/(x − 1) = 2. First, x ≠ 1. Cancelling gives x + 1 = 2, but x ≠ 1 remains. Its only candidate, 1, is excluded, so the original equation has no solutions. Substitution would produce 0/0, not 2. Rational equations, OpenStax §2.2.
Squaring erases a sign the equation needs
A radical equation here contains a variable under a root sign. In √(x + 2) = x, the radical’s domain is x ≥ − 2. But the left side is nonnegative, so equality additionally requires x ≥ 0.
Squaring gives x + 2 = (x)^(2), or (x − 2)(x + 1) = 0. Solving this quadratic without the sign restriction yields 2 and −1. At 2 the original says √(4) = 2; at −1, √(1) = 1 ≠ − 1. Only x = 2 remains.
Original equation
y = √(x + 2); y = x
After squaring
y = x + 2; y = x²
| Candidate x | Check √(x+2)=x | Decision |
|---|---|---|
| −1 | 1 ≠ −1 | Reject |
| 2 | 2 = 2 | Keep |
The number −1 is allowed in the radical’s domain, but does not solve the equation. It is an extraneous candidate after squaring, not a forbidden radicand input. If we retain x ≥ 0 from the start, the transition to x + 2 = (x)^(2) is equivalent under that condition; the extra candidate is rejected even before substitution. Radical equations and extraneous solutions, OpenStax §2.6.
A substitution has a range and a return step
For x + √(x) − 6 = 0, the domain is x ≥ 0. Set t = √(x), hence t ≥ 0 and x = (t)^(2). We obtain (t)^(2) + t − 6 = 0, or (t − 2)(t + 3) = 0. The value t = − 3 cannot be a principal square root, leaving t = 2, x = 4. Check: 4 + 2 − 6 = 0.
Taking both t values and simply squaring would wrongly add x = 9. The original would give 9 + 3 − 6 = 6, not zero. A restriction on the new variable is as necessary as one on the original.
Nonlinear systems: a graph answers its own question
The second panel can be read as the separate system y = (x)^(2), y = x + 2. It is nonlinear because one equation contains a squared variable. Substitute y = x + 2 into the first to obtain (x)^(2) = x + 2, giving x = − 1 or 2. Recovering y yields ( − 1,1) and (2,4). Both satisfy this system.
Here −1 is not extraneous: the system truly has two intersections. But it describes equality of squares; the original √(x + 2) = x also required nonnegative x. A graphical check helps when we label exactly which equation or system is drawn. Substitution into both conditions confirms exact coordinates. Nonlinear systems, OpenStax §11.3.
Branch · CompletenessA link to equivalent transformations.
Checking cannot recover a lost solution
Dividing x(x − 1) = 0 by x leaves only 1. Substitution confirms it but will not find the omitted 0. Preserve every case and restriction first, then use checking to reject extra candidates. Candidate verification and proof of completeness are both parts of solving.
A list of numbers becomes an answer after returning to the original denominators, signs and variable relationships. One useful question controls each step: can we return from the new equation to the old one on the stated domain?