BSEH Secondary Mathematics - Basic common mistakes
BSEH Secondary Mathematics - Basic · Understand the traps and how to avoid them.
When rationalising a denominator such as 1/(√7−√5), the candidate multiplies numerator and denominator by (√7−√5) instead of the conjugate (√7+√5), so the denominator becomes 7−5=2 with a sign error and the answer comes out as (√7−√5)/2 rather than (√7+√5)/2.
The procedure is memorised as 'multiply by the same surd' rather than as 'multiply by the conjugate to create a difference of squares'. Under time pressure the candidate pattern-matches to the visible expression instead of reasoning about which factor removes the surd.
What to do instead
Before multiplying, write the target on the answer line: denominator must become a²−b². Practise by first stating the conjugate aloud, then multiplying. Verify by substituting a decimal approximation (e.g. √7≈2.646, √5≈2.236) into original and final expressions; a sign error shows immediately as a mismatch.
The candidate states that every terminating decimal is rational but also claims √2 = 1.414 is rational 'because it terminates', confusing the rounded decimal display with the exact value.
Calculators display √2 to a finite number of places, and the candidate has never separated the idea of a decimal representation from the number itself. The irrationality proof by contradiction is often skipped in revision.
What to do instead
Learn one full proof that √2 is irrational by contradiction (assume √2 = p/q in lowest terms, derive that p and q are both even). Then classify a list of numbers by exact reasoning, not by calculator output, and note that any non-terminating non-repeating decimal is irrational.
Solving a pair of linear equations by elimination, the candidate subtracts the equations but forgets to change signs across the whole equation, e.g. subtracting 3x+2y=12 from 5x+2y=16 and writing 2x=28 instead of 2x=4.
Sign distribution across a multi-term expression is a fragile skill, and the candidate performs the subtraction term by term mentally without writing the intermediate line.
What to do instead
Always write the subtraction explicitly as a full line with brackets: (5x+2y)−(3x+2y)=16−12, then expand. After finding x and y, substitute both back into the equation that was not used for elimination; this catches nearly every sign slip.
For a quadratic with roots α and β, the candidate computes α+β = −b/a but then writes the new equation with roots 1/α and 1/β as x² − (1/α+1/β)x + 1/(αβ) = 0 and substitutes b and c directly without converting, producing a wrong constant term.
The relationship between coefficients and roots is memorised only for the standard case; the candidate does not rebuild sums and products for the transformed roots.
What to do instead
Use the identity route every time: 1/α+1/β = (α+β)/(αβ) and (1/α)(1/β) = 1/(αβ). Compute these two numbers first, then write x² − (sum)x + (product) = 0. Check with a concrete example such as roots 2 and 3, whose reciprocals should give 6x²−5x+1=0.
In an arithmetic progression the candidate uses aₙ = a + nd instead of aₙ = a + (n−1)d, so the 10th term is computed with 10d rather than 9d, and the sum formula is similarly off by one term.
The formula is recalled as a shape rather than derived; the candidate never checks that the first term corresponds to n=1.
What to do instead
Derive it once: from a to the nth term there are (n−1) steps of d. Then always test the formula on n=1 (should return a) and n=2 (should return a+d) before using it in a problem.
Finding the point dividing a segment in a given ratio, the candidate substitutes coordinates into the section formula without checking whether the ratio is internal or external, and mixes the order of the ratio (using m₁ with the second point).
The formula is remembered as a single template and the meaning of m₁ : m₂ relative to the endpoints is not fixed in the candidate's mind.
What to do instead
Label the endpoints A(x₁,y₁) and B(x₂,y₂) on a sketch and write the ratio as AP:PB before touching the formula. For external division use (mx₂−nx₁)/(m−n). Test with the midpoint ratio 1:1, which must return the average of the coordinates.
Proving two triangles similar, the candidate writes 'AA similarity' after naming one pair of equal angles and one pair of equal sides, treating a side condition as if it established an angle.
The similarity criteria (AA, SSS, SAS) are listed as labels without the candidate internalising what each letter requires; equal sides feel like strong evidence.
What to do instead
For each criterion write the required ingredients next to it: AA needs two angle pairs; SAS needs two side pairs in the same ratio with the included angle equal; SSS needs all three side ratios equal. In every proof, state each equality with its justification (vertically opposite, alternate angles, common side) before naming the criterion.
Using the tangent–secant or two-tangent theorem, the candidate assumes a line is tangent because it touches the drawn circle, without the right angle at the point of contact, and applies the tangent-length equality to a chord.
Diagrams in textbooks look tangent, and the candidate reasons from the picture rather than from the stated hypothesis.
What to do instead
Before applying any tangent theorem, write 'radius ⊥ tangent at point of contact' as a checked line. If the right angle is not given or provable, the line is a secant and a different theorem applies. Redraw the figure with an obviously non-tangent line to break the visual habit.
Computing the angle of elevation, the candidate sets tan θ = horizontal distance / height instead of height / horizontal distance, or uses the hypotenuse where the horizontal distance belongs.
The candidate selects the ratio by habit rather than by identifying which sides are given relative to the angle.
What to do instead
Draw the right triangle, mark the angle at the observer, and label each side as opposite, adjacent or hypotenuse relative to that angle before choosing a ratio. Write SOH-CAH-TOA above the working and tick the ratio chosen. Verify that an angle of elevation for a tall object comes out large.
In a frustum or combined-solids problem, the candidate adds the surface areas of the two component solids including the faces that are glued together, or uses the full cone slant height for the frustum's curved surface.
The formula list is applied solid by solid, and the candidate does not visualise which surfaces remain exposed after joining.
What to do instead
Sketch the assembled solid and shade the exposed surfaces only; cross out the hidden circular face. For a frustum use slant height l = √(h²+(R−r)²) and curved area π(R+r)l. As a check, the total surface area of the joined solid must be less than the sum of the separate solids.
Finding the median of grouped data, the candidate uses n/2 as the position and reads the median class from cumulative frequency without locating the class that first exceeds n/2, or omits the class width in the interpolation formula.
The formula median = l + ((n/2 − cf)/f) × h is memorised as symbols without the meaning of cf (cumulative frequency of the class before the median class) and h (class width).
What to do instead
Build the cumulative frequency column first, mark where n/2 falls, and circle the median class. Label l, cf, f and h on the table itself before substituting. Sanity-check that the answer lies inside the median class interval.
In probability, the candidate adds probabilities of two events that can occur together (e.g. drawing a king or a heart) and gets 4/52 + 13/52 = 17/52 instead of subtracting the overlap.
The addition rule is remembered without the inclusion–exclusion correction, and the candidate does not check whether the events are mutually exclusive.
What to do instead
Before adding, ask whether both events can happen at once. If yes, use P(A∪B) = P(A) + P(B) − P(A∩B) and identify the overlap explicitly (the king of hearts). Check that no probability exceeds 1 and that the answer is at least the larger of the two individual probabilities.