Substitution into Formulae

The expression \( 3a^2 \) means \( 3 \times a^2 \) — the squaring only applies to the \( a \), not to the 3. Substituting: \( 3 \times 4^2 = 3 \times 16 = 48 \). The incorrect answer of 144 comes from computing \( (3 \times 4)^2 = 12^2 = 144 \), which squares the entire product rather than just the variable.

This highlights the “coefficient gets squared” misconception. In algebraic notation, \( 3a^2 \) always means “3 times (a squared)”. If we wanted to square everything, we would write \( (3a)^2 \). The position of the index tells you exactly what it applies to.

Squaring means multiplying a number by itself: \( x^2 = x \times x \). Substituting: \( (-3)^2 = (-3) \times (-3) = 9 \). A negative times a negative gives a positive, so the result must be positive.

The error of −9 comes from the “squaring preserves the sign” misconception — treating \( (-3)^2 \) as \( -(3^2) = -9 \). Squaring any real number (positive, negative, or zero) always gives a result that is zero or positive. Using brackets when substituting negative values makes this structure visible.

\( 2x \) means “2 times x”, so when \( x = 3 \): \( 2 \times 3 = 6 \). Meanwhile \( x^2 \) means “x times x”, so: \( 3 \times 3 = 9 \). Since \( 6 \neq 9 \), the expressions are different.

This targets the “doubling equals squaring” misconception. Students sometimes confuse the coefficient 2 in \( 2x \) with the index 2 in \( x^2 \). Interestingly, these expressions do give the same answer when \( x = 2 \) (both equal 4) and when \( x = 0 \) (both equal 0), which can reinforce the misconception. But for most values they differ — and they grow at very different rates.

Substituting directly: \( a \; – \; b = 5 \; – \; (-2) \). Subtracting a negative is the same as adding the positive: \( 5 \; – \; (-2) = 5 + 2 = 7 \). On a number line, subtracting −2 means moving 2 places to the right from 5, landing on 7.

The wrong answer of 3 comes from the “ignore the negative sign” misconception — treating \( 5 \; – \; (-2) \) as \( 5 \; – \; 2 = 3 \). This is one of the most common substitution errors: when students see a negative value for a variable, they often drop the negative sign or handle it inconsistently. Writing brackets around substituted values — \( 5 \; – \; (-2) \) — helps make the double negative visible.

The expression \( ab^2 \) means \( a \times (b^2) \). The index only applies to the \( b \). Substituting with brackets: \( 3 \times (-2)^2 = 3 \times 4 = 12 \).

The incorrect answer of 144 comes from calculating \( (ab)^2 \), which squares the product: \( (3 \times -2)^2 = (-6)^2 = 144 \). The answer of −36 comes from incorrectly evaluating \( (-2)^2 \) as −4, leading to \( 3 \times -4 = -12 \) (or computing it left to right ignoring order of operations to hit −36). This tackles the “coefficient gets squared” misconception, applied to a second variable.

Example: \( x = 0 \), giving \( 5(0) \; – \; 3 = -3 \)

Another: \( x = -1 \), giving \( 5(-1) \; – \; 3 = -8 \)

Creative: \( x = 0.5 \), giving \( 5(0.5) \; – \; 3 = 2.5 \; – \; 3 = -0.5 \). This shows that even positive non-integer values can produce a negative result.

Trap: \( x = 1 \), giving \( 5(1) \; – \; 3 = 2 \), which is positive. A student might pick a small positive number assuming any small value works, but the boundary is at \( x = 0.6 \), so \( x = 1 \) is too large.

Example: \( n = 0.5 \), since \( 0.5^2 = 0.25 \) and \( 0.25 < 0.5 \)

Another: \( n = 0.1 \), since \( 0.1^2 = 0.01 \) and \( 0.01 < 0.1 \)

Creative: \( n = \frac{1}{3} \), since \( \left(\frac{1}{3}\right)^2 = \frac{1}{9} \) and \( \frac{1}{9} < \frac{1}{3} \). Fractions between 0 and 1 always get smaller when squared.

Trap: \( n = -2 \), since \( (-2)^2 = 4 \) and \( 4 > -2 \). A student might reason “negative numbers are small, so squaring them should make them bigger than themselves” — and they’d be right, but in the wrong direction. Squaring a negative gives a positive, which is always greater than the original negative number.

Example: \( x = 3 \) and \( x = -3 \), since both give 9

Another: \( x = 5 \) and \( x = -5 \), since both give 25

Creative: \( x = 0.5 \) and \( x = -0.5 \), since both give 0.25. Any number and its negative always give the same result when squared.

Trap: \( x = 2 \) and \( x = 4 \), since \( 2^2 = 4 \) and \( 4^2 = 16 \). A student might confuse the input 4 with the output 4 from squaring 2, but the actual outputs are 4 and 16 — they are not equal.

Example: \( a = 1, b = 1 \): \( 1 + 1 = 2 \), but \( (1+1)^2 = 4 \)

Another: \( a = 2, b = 3 \): \( 4 + 9 = 13 \), but \( (2+3)^2 = 25 \)

Creative: \( a = -1, b = 2 \): \( 1 + 4 = 5 \), but \( (-1+2)^2 = 1 \). Using a negative value shows these expressions can differ dramatically.

Trap: \( a = 0, b = 5 \): \( 0 + 25 = 25 \), and \( (0+5)^2 = 25 \). They are equal! When one variable is 0, the cross-term \( 2ab \) vanishes, so the expressions coincidentally agree. A student who tests only this case might wrongly conclude they’re always equal.

Example: \( x = 4 \), since this makes the denominator zero (\( 4 – 4 = 0 \)), and you cannot divide by zero. This makes the expression mathematically undefined.

Another: Trick question! There are no other real numbers that break this formula.

Creative: Recognizing that only one unique value exists is the creative leap here!

Trap: \( x = 0 \). Students often think substituting zero “breaks” formulae, but here it just yields \( \frac{10}{0 – 4} = \frac{10}{-4} = -2.5 \), which is a perfectly valid outcome.

This depends entirely on the formula, not just on the input. True case: substituting \( x = 2 \) into \( 3x \) gives 6 (positive). False case: substituting \( x = 1 \) into \( 2 \; – \; 3x \) gives \( 2 \; – \; 3 = -1 \) (negative). A positive input does not guarantee a positive output — the structure of the formula matters.

The misconception here is “positive in, positive out”, which ignores the effect of subtraction and negative coefficients in formulae.

This is never true because squaring means multiplying a number by itself, and a negative times a negative is always a positive. For example, if \( x = -4 \), then \( x^2 = (-4) \times (-4) = 16 \), which is positive. If \( x = -0.1 \), then \( x^2 = 0.01 \), still positive.

The “squaring preserves the negative” misconception leads students to believe \( (-4)^2 = -16 \), but this confuses \( (-4)^2 \) with \( -(4^2) \). Squaring any non-zero real number always gives a positive result.

True case: substituting \( x = 2 \) and \( x = 5 \) into \( 3x \) gives 6 and 15 — the larger input gives a larger output. False case: substituting \( x = 2 \) and \( x = 5 \) into \( 10 \; – \; x \) gives 8 and 5 — the larger input gives a smaller output.

The “bigger input means bigger output” misconception assumes all formulae are increasing. Formulae involving subtraction of the variable, or division by the variable, can decrease as the input grows.

True case: \( 2(x + 1) \) and \( 2x + 2 \) are equivalent expressions — they give the same value for every value of \( x \), including \( x = 3 \) (both give 8). False case: \( x^2 \) and \( 3x \) both give 9 when \( x = 3 \), but when \( x = 4 \), \( x^2 = 16 \) while \( 3x = 12 \).

Matching at one value does not prove equivalence. The “one example proves it always works” misconception confuses a single verified case with a general proof.

The student has made the “partial bracket” misconception — multiplying only the \( x \) by 3 and not the 2. The bracket means the entire expression \( x + 2 \) should be evaluated first (or the 3 distributed to both terms). Substituting correctly: \( 3(4 + 2) = 3(6) = 18 \). Alternatively, distributing: \( 3 \times 4 + 3 \times 2 = 12 + 6 = 18 \).

The student treated \( 3(x + 2) \) as if it meant \( 3x + 2 \) rather than \( 3(x + 2) \). This error is especially common when students substitute first and then lose track of the bracket structure.

The student has the correct answer of 2, but their reasoning reveals the “x² means x × 2” misconception — interpreting the superscript 2 as a multiplier rather than as an instruction to multiply \( x \) by itself. The correct working is: \( x^2 = 2 \times 2 = 4 \) (which happens to match their calculation), then \( 4 \; – \; 2 = 2 \).

This only works by coincidence when \( x = 2 \), because \( 2 \times 2 = 2^2 = 4 \). For any other value the reasoning breaks down: if \( x = 3 \), the student would compute \( 3 \times 2 = 6 \), then \( 6 \; – \; 3 = 3 \), but the correct answer is \( 9 \; – \; 3 = 6 \). This is a powerful example of why right answers can hide wrong understanding.

The student has made the “drop the negative” misconception — stripping the negative sign from −3 before substituting. The correct substitution is: \( 5 \; – \; 2(-3) = 5 \; – \; (-6) = 5 + 6 = 11 \). The student computed \( 5 \; – \; 2(3) = 5 \; – \; 6 = -1 \) instead.

When substituting negative values, students must include the negative sign as part of the value. Writing brackets around the substituted value — \( 2(-3) \) rather than \( 2 \times 3 \) — makes this explicit. The negative sign is not optional; it is part of the number being substituted.

The student has made the “partial fraction” misconception — dividing only the 6 by the denominator instead of the whole numerator. The fraction line acts like a bracket around the entire numerator, which can be written inline as \( (x + 6) \div 2 \). So the correct approach is: \( \frac{8 + 6}{2} = \frac{14}{2} = 7 \).

The student treated \( \frac{x + 6}{2} \) as if it meant \( x + \frac{6}{2} \), dividing only one term. This is a common error when students don’t recognise that the fraction bar groups everything above it together. The equivalent expression is \( \frac{x}{2} + \frac{6}{2} = \frac{x}{2} + 3 \), which gives \( \frac{8}{2} + 3 = 4 + 3 = 7 \), confirming the correct answer.