Substituting into Formulae

Activities to try

1. 1

   Rebuild-the-expression substitution with a sliding value

   Why it works: Substitution means replacing the letter with its value while preserving the encoded operations; brackets protect negatives so (−2)² = +4, and a² (=a×a) behaves differently from 2a (=2×a).

   1. Take the expression 3a² − 2a and first rewrite the implicit operations explicitly: 3×a×a − 2×a, or 3×(a)² − 2×(a).
   2. Choose a value for a and drop it inside the brackets so the expression becomes 3×( )² − 2×( ), e.g. 3×(5)² − 2×(5).
   3. Evaluate in order: resolve the power first, then the multiplications, then the subtraction. For a=5: 3×25 − 10 = 75 − 10 = 65.
   4. Repeat the same expression for a sequence of values — try a = 3, then a = 0, then a = −2, then a = 1/2 — writing out every step each time.
   5. For each value, note alongside what the two parts (a² and 2a) came to separately, and watch how a² stays non-negative while 2a goes negative as a crosses zero.
2. 2

   Contrast a² against 2a on the same value

   Why it works: a² means a×a, not a×2; the exponent is not a coefficient, so the two expressions generally give very different values (coinciding only at a=0 and a=2).

   1. Pick a value, e.g. a=5, and evaluate a² and 2a separately: a² = 5×5 = 25, 2a = 2×5 = 10.
   2. Draw a 5-by-5 square for a² next to a bar of length 10 for 2a, and compare their sizes.
   3. Repeat for a=0 (both give 0) and for a=3, recording how the two results diverge.
   4. State in your own words why a² and 2a are generally not the same, and identify when they do coincide.
3. 3

   Negative and fractional values that expose the rules

   Why it works: Brackets around a substituted value preserve its sign under powers, so (−2)² = 4; −a² and (−a)² are different; and special values like 0 and fractions must still be evaluated term by term with correct order of operations.

   1. Evaluate 3a² − 2a at a = −2, substituting inside brackets: 3×(−2)² − 2×(−2) = 3×4 + 4 = 12 + 4 = 16.
   2. Evaluate −a² at a = 3 and at a = −3, substituting inside brackets (−(3)² and −(−3)²), and check that BOTH give −9; then evaluate (−a)² at the same values, (−3)² and (3)², which both give +9 — confirming −a² differs from (−a)².
   3. Evaluate the expression at a = 1/2, keeping the fraction inside the bracket for the square and ordering the operations correctly.
   4. Try a = 0 and check each term explicitly, including any constant, instead of assuming everything vanishes.
   5. Optionally use a number line to place each negative substituted value and keep track of its sign before applying the power.