Chapter 5.4 — Exercise 5.3 — Completing the Square
Solving quadratic equations by completing the square. This is Lesson 4 of 5 in Chapter 5: Quadratic Equations.
A Method That Never Gets Stuck
Factorisation is fast when a quadratic splits cleanly, but plenty of equations refuse to split into neat integer factors at all, however patiently you search for the right pair. Exercise 5.3 introduces completing the square — a method that works on every quadratic without exception — and uses it to derive a single formula that solves any of them directly, on the first attempt.
Deriving the Quadratic Formula
Starting from the general equation ax² + bx + c = 0, five steps turn it into a formula for x purely in terms of a, b, and c:
- Step 1: multiply through by 4a: 4a²x² + 4abx + 4ac = 0.
- Step 2: add b² to both sides: 4a²x² + 4abx + b² = b² − 4ac.
- Step 3: the left side is now a perfect square: (2ax + b)² = b² − 4ac.
- Step 4: take the square root of both sides: 2ax + b = ±√(b² − 4ac).
- Step 5: solve for x: x = [−b ± √(b² − 4ac)] / 2a.
x = [−b ± √(b² − 4ac)] / 2a — valid whenever b² − 4ac ≥ 0This one formula replaces the trial-and-error search for factor pairs entirely — plug in a, b, and c, and both roots fall out directly, with no guessing involved at any step, and no dependence on the coefficients happening to be small, whole, or otherwise convenient.
It's worth walking back through why each step is valid, not just accepting the final formula on faith. Multiplying by 4a (Step 1) is reversible as long as a≠0, which the definition of a quadratic equation already guarantees. Adding b² to both sides (Step 2) doesn't change the equation's truth, only its shape. And the algebra behind Step 3 is the same perfect-square pattern seen throughout this exercise: 4a²x² + 4abx + b² really is (2ax)² + 2(2ax)(b) + b², which is exactly the expansion of (2ax+b)² read backward.
Four Equations, Solved Both Ways
The exercise solves the same four equations twice — first by completing the square from scratch, then by substituting directly into the formula just derived — confirming both methods land on exactly identical roots every single time.
| Equation | a, b, c | Roots |
|---|---|---|
| 2x² + x − 4 = 0 | 2, 1, −4 | (−1 ± √33) / 4 |
| 4x² + 4√3x + 3 = 0 | 4, 4√3, 3 | −√3/2, −√3/2 (repeated) |
| 5x² − 7x − 6 = 0 | 5, −7, −6 | 2, −3/5 |
| x² + 6x + 5 = 0 | 1, 6, 5 | −1, −5 |
The second row is worth a closer look: b² − 4ac works out to exactly 0 here (48 − 48), so the ± in the formula makes no difference at all — both roots land on the same value, −√3/2, the algebraic signature of a repeated root already seen from factorisation in the previous exercise.
Two Equations That Need Rearranging First
Not every equation arrives in ax² + bx + c = 0 form — some need clearing fractions before the formula can be applied at all.
- x − 1/x = 3 (x ≠ 0) — multiplying through by x gives x² − 3x − 1 = 0, with roots (3 ± √13)/2.
- 1/(x+4) − 1/(x−7) = 11/30 (x ≠ −4, 7) — combining the fractions and cross-multiplying gives x² − 3x + 2 = 0, with roots 2 and 1.
Rehman's Age, Recovered From a Fraction Sum
The sum of the reciprocals of Rehman's age 3 years ago and 5 years from now is 1/3. Naming his present age x:
1/(x−3) + 1/(x+5) = 1/3 ⟹ x² − 4x − 21 = 0The quadratic formula gives x = 7 or x = −3 — and since an age can't be negative, Rehman's present age is 7 years. Checking: 3 years ago he was 4, in 5 years he'll be 12, and 1/4 + 1/12 = 3/12 + 1/12 = 4/12 = 1/3 ✓.
Six More Word Problems
The same formula-and-check pattern — form the equation, solve with the quadratic formula, reject any root that doesn't fit the real situation — carries through the rest of the exercise:
| Situation | Answer |
|---|---|
| Moulika's Maths+English marks total 30; +2 and −3 respectively multiply to 210 | 13 & 17, or 12 & 18 (both valid) |
| Rectangular field: diagonal 60 m more than shorter side, longer side 30 m more | Shorter 90 m, longer 120 m |
| Difference of squares of two numbers is 180; smaller² = 8 × larger | 18 and 12 |
| Train travels 360 km; 5 km/h faster would save 1 hour | 40 km/h |
| Two taps fill a tank in 9⅜ hours together; larger tap 10 hours faster alone | Smaller tap 25 hrs, larger tap 15 hrs |
| Express train 11 km/h faster than passenger, 1 hour less for 132 km | Passenger 33 km/h, express 44 km/h |
The water-taps problem produces a second algebraic root, x = 15/4 hours, that gets rejected for a subtler reason than "negative" — the larger tap's time would then be x−10, a negative number, even though x itself stayed positive. Checking a root sometimes means testing every quantity built from it, not just the single variable that was originally solved for in the first place.
A Projectile and a Polygon
An object thrown upward from a 12 m building follows S = 96 + 80t − 16t², reaching the ground when S = 0:
−16t² + 80t + 96 = 0 ⟹ t² − 5t − 6 = 0 ⟹ (t−6)(t+1) = 0Since a negative time makes no physical sense here, the object reaches the ground after exactly 6 seconds. The final problem asks how many sides a polygon with 65 diagonals has, using the formula n(n−3)/2 for the diagonal count: solving n² − 3n − 130 = 0 gives n = 13 (the negative root is rejected), so a 13-sided polygon has exactly 65 diagonals. But asking the same question for 50 diagonals gives n² − 3n − 100 = 0, whose roots are (3 ± √409)/2 — and since √409 isn't a whole number, n itself never comes out to a whole number, meaning no polygon with 50 diagonals can exist at all.
That last result is genuinely worth sitting with for a moment: it isn't that the equation was solved incorrectly anywhere, it's that a perfectly valid equation can have real roots that simply don't correspond to anything physically meaningful — a polygon must always have a whole number of sides, and this particular equation's roots simply aren't whole numbers, full stop.
Knowing Before Solving
Every single problem above eventually asked "does a real answer exist here, and does it actually make sense?" — a question answered only after fully solving the equation. Exercise 5.4 shows how to answer the first half of that question — whether real roots exist at all — by inspecting b² − 4ac alone, before doing any of the actual solving work this exercise required each time. For the factorisation method this exercise builds an alternative to, see Exercise 5.2, or revisit the standard-form and root definitions laid out in the chapter introduction.