NY Regents Algebra I common mistakes
NY Regents Algebra I · Understand the traps and how to avoid them.
When simplifying a radical like √72, the candidate writes √72 = 6√2 but then also writes √(a+b) = √a + √b for expressions such as √(9+16), reporting 7 instead of 5.
The distributive-looking pattern feels familiar, so candidates over-extend it to radicals. They memorize 'pull out perfect squares' as a procedure without checking it against the definition of a square root, so the additive version slips through.
What to do instead
Have them test every radical claim numerically first: √25 = 5, but √9 + √16 = 7, so the split is false. Rebuild the rule from the definition: √(ab) = √a·√b only when the operation inside is multiplication. Then drill a mixed set where some items are products and some are sums, and require a one-line justification ('product rule applies' vs 'sum inside, cannot split') before each answer.
The candidate multiplies (x+3)² and writes x² + 9, dropping the middle term.
Squaring is treated as 'apply the exponent to each piece', which works for (3x)² but not for a binomial. The candidate has no mental image of the area model, so the cross terms have nowhere to come from.
What to do instead
Draw the 2-by-2 area square with side (x+3): four regions, x², 3x, 3x, 9, giving x² + 6x + 9. Do the same for (x−5)² and (2x+1)². Then have them expand symbolically with distribution and compare to the picture. On any multiple-choice item, they should plug x = 2 into both the original and their answer to catch a mismatch.
Solving 2x + 3 < 11, the candidate subtracts and divides correctly but writes x > 4, or when solving −3x ≥ 12 writes x ≥ −4.
The flip rule is memorized as a slogan ('flip when negative') without a model of why. Candidates also confuse 'the coefficient is negative' with 'the operation was division by a negative', so they flip at the wrong step or not at all.
What to do instead
Anchor the rule in a number-line test: start from 4 < 6, multiply both sides by −1, and see that −4 > −6. Then have them solve every inequality twice, once algebraically and once by testing a value on each side of the boundary. Require them to state which step required the flip and why.
Given a system of two linear equations, the candidate solves for one variable, substitutes back into the same equation they started from, and declares a unique solution from a single equation.
The substitution procedure is memorized as a sequence of moves without the purpose of the second equation. Candidates also skip checking whether the lines are parallel or identical, so 'no solution' and 'infinitely many' items get a fabricated point.
What to do instead
Force a two-column layout: solve for y from equation (1), then substitute into equation (2) specifically, and label which equation is being used at each step. Before solving, compute both slopes; if equal, stop and classify (same intercept = infinitely many, different intercept = none). Verify the final point in both original equations and show the check.
Asked for the vertex of y = x² − 6x + 5, the candidate reads the vertex as (0, 5) from the constant term, or computes x = −b/a instead of x = −b/(2a).
The constant term is visually salient and is confused with the y-intercept's role in the vertex. The formula is recalled partially, and candidates do not connect the vertex to the axis of symmetry or to completing the square.
What to do instead
Have them complete the square once by hand: x² − 6x + 5 = (x − 3)² − 4, so the vertex is (3, −4). Then show that −b/(2a) = 6/2 = 3 matches, and that (0, 5) is the y-intercept, a different feature. Require every quadratic answer to include a quick sketch with vertex, axis of symmetry, and intercepts labeled.
When a function is given as f(x) = 2x − 1, the candidate computes f(a + 1) as 2a − 1 + 1, treating f as multiplication by a constant rather than as an instruction to substitute.
The notation f(x) looks like a product, and candidates have not internalized function notation as 'input goes here'. Nested or shifted inputs expose this immediately.
What to do instead
Use an input-output machine diagram: whatever is written inside the parentheses is the input and must replace every x. Work f(3), f(0), f(−2), then f(a), then f(a+1) = 2(a+1) − 1 = 2a + 1, showing each substitution explicitly. Follow with composition f(g(x)) to cement that the whole inner expression is the input.
The candidate computes the slope between (2, 5) and (6, 13) as (6 − 2)/(13 − 5) = 1/2, inverting the rise-over-run ratio.
Under time pressure the subtraction order feels arbitrary, and candidates do not consistently pair the y-values together and the x-values together. They also skip labeling the points, so the pairing is lost.
What to do instead
Require labeling each point (x₁, y₁) and (x₂, y₂) before substituting, and write the formula with the labels visible: (y₂ − y₁)/(x₂ − x₁). Sanity-check with a sketch: from (2,5) to (6,13) the line rises 8 while running 4, so the slope must exceed 1. Any answer below 1 should trigger a recheck.
The candidate writes the equation of a line through (4, 7) with slope 3 as y = 3x + 7, ignoring that the y-intercept is not the given y-value.
The slope-intercept form is over-applied; candidates see a point and a slope and reach for y = mx + b without solving for b. The point (4, 7) is mistaken for (0, 7).
What to do instead
Have them substitute the point into y = mx + b: 7 = 3(4) + b, so b = −5, giving y = 3x − 5. Then verify by plugging x = 4 back in. Pair this with point-slope form y − y₁ = m(x − x₁) and show both routes give the same line. Insist on the final check that the given point satisfies the final equation.
In a circle problem, the candidate uses the diameter in place of the radius, or computes area with 2πr and circumference with πr².
The two formulas look similar and are often memorized as a pair of letters rather than as quantities with units. Candidates also grab whichever number is printed on the diagram without checking whether it spans the center.
What to do instead
Attach units to every formula: circumference is a length (π·d, one dimension), area is square units (π·r², two dimensions). Have them label radius and diameter on a sketch before computing, and check that an area answer is in square units and larger than the circumference numerically in typical problems. Drill a mixed set where the diameter is given.
Given a set of data, the candidate reports the mean as the 'typical' value without noticing an outlier, or computes the median of an even-sized list by picking the middle item instead of averaging the two middle items.
The mean is the most-practiced measure, so it becomes the default. Candidates also treat 'median' as a positional lookup and stop at the first middle value, forgetting that an even count has two.
What to do instead
Have them sort the list first and write the count n at the top. If n is even, circle both middle values and average them. Then compare mean and median for the same data set with and without an extreme value, and ask which better describes a typical case. On any 'which measure' item, require a one-sentence justification referencing the outlier or the shape of the data.