CLEP College Algebra common mistakes
CLEP College Algebra · Understand the traps and how to avoid them.
When simplifying a rational expression such as (x² − 9)/(x − 3), the candidate cancels the x² against the x and the 9 against the 3, writing x − 3 instead of x + 3.
The candidate treats terms inside a sum as if they were factors, because the visual pattern of repeated symbols suggests cancellation. Cancellation is only legal between factors of a product, and the numerator is a difference of squares that must be factored first.
What to do instead
Before cancelling anything, factor the numerator and denominator completely and rewrite the expression as a product of factors. Then cross out only identical factors, one at a time, and verify by substituting a number such as x = 4 into both the original and simplified forms to confirm they agree.
When adding rational expressions with unlike denominators, the candidate adds numerators and adds denominators, writing 1/x + 1/y = 2/(x + y).
Fraction addition rules learned with numbers are misapplied to symbolic fractions; the candidate remembers 'add the tops, add the bottoms' from a confused memory of multiplication of fractions.
What to do instead
Rewrite every sum over a single common denominator, multiplying each numerator by the missing factor, then combine only the numerators. Check with numeric values: 1/2 + 1/3 = 5/6, not 2/5, so the symbolic rule must mirror the numeric one.
When solving an equation such as 3(x − 2) = 12, the candidate distributes over only the first term, producing 3x − 2 = 12.
Under time pressure the candidate skips the distribution step and mentally 'moves' the 3 only to the nearest term, a habit formed from simpler equations where multiplication applied to a single term.
What to do instead
Write the distribution explicitly on paper before doing any other operation, drawing arrows from the multiplier to each term inside the parentheses. Then verify by substituting the solution back into the original equation.
When solving an inequality such as −2x > 6, the candidate divides by −2 and writes x > −3, forgetting to reverse the inequality sign.
The candidate treats inequalities as equations, since the mechanical steps are otherwise identical; the sign-reversal rule is easy to forget because it has no counterpart in equation solving and rarely appears in drill problems.
What to do instead
Adopt the habit of marking the moment you multiply or divide by a negative with an explicit note, then flip the symbol. As a safety check, test a value: if x = −4, then −2(−4) = 8 > 6 is true, so x > −3 cannot be correct.
When solving an absolute-value equation such as |x − 3| = 7, the candidate writes only x − 3 = 7 and reports x = 10, dropping the negative case.
The candidate converts the absolute value bars directly into parentheses, forgetting that the expression inside can equal either the positive or the negative of the right side.
What to do instead
Always split into two equations, x − 3 = 7 and x − 3 = −7, and solve both. If the right side is negative, state immediately that there is no solution, since an absolute value cannot equal a negative number.
After solving a radical or rational equation, the candidate reports every algebraic solution without checking, including extraneous roots that do not satisfy the original equation.
Squaring both sides or clearing denominators can introduce solutions that were never valid; the candidate treats the derived equation as equivalent to the original when it is only a consequence of it.
What to do instead
Treat every solution of a squared or cleared equation as a candidate and substitute it into the original equation. Reject any candidate that makes a denominator zero or a radicand negative, and say so explicitly in the answer.
When asked for the domain of f(x) = √(x − 4)/(x − 6), the candidate gives only x ≥ 4 or only x ≠ 6, missing one of the two restrictions.
The candidate scans for one familiar restriction (a square root or a denominator) and stops, rather than auditing every operation in the expression for domain constraints.
What to do instead
Run a checklist over the expression: denominators cannot be zero, even roots need nonnegative radicands, and logarithms need positive arguments. Combine all restrictions with intersection, then test boundary values numerically to confirm the interval.
When finding the inverse of a function, the candidate swaps x and y but then solves for y incorrectly, or claims the inverse of f(x) = x² is x² without restricting the domain.
The swap step is memorable and gets done, but the algebra that follows is rushed, and the candidate forgets that a function must be one-to-one to have an inverse function on its full domain.
What to do instead
After swapping, isolate y step by step and verify by composing: f(f⁻¹(x)) should return x. For even powers, restrict the domain explicitly (for example x ≥ 0) before inverting, and state the restriction in the answer.
When composing functions, the candidate computes f(g(x)) as f(x)·g(x), multiplying the two functions instead of substituting one into the other.
The notation f(g(x)) looks like juxtaposition, which elsewhere means multiplication, and the candidate has not internalized that the inner expression replaces every x in the outer function.
What to do instead
Rewrite the outer function with a blank in place of every x, then fill the blank with the entire inner expression, parentheses included. Evaluate at a test number such as x = 2 to confirm the composition differs from the product.
When working with complex numbers, the candidate writes √(−9)·√(−4) = √36 = 6 instead of −6, or reports i² as 1.
The candidate applies the product rule for radicals, which holds only for nonnegative radicands, and loses track of the definition i² = −1 during multi-step simplification.
What to do instead
Convert every negative radicand to i times the positive root before multiplying, so √(−9)·√(−4) = 3i · 2i = 6i² = −6. Write i² = −1 explicitly at each step until it becomes automatic, and never combine radicals with negative radicands under a single root sign.