Completing the square on ax² + bx + c = 0 with letters instead of numbers produces one formula that solves every quadratic. Deriving it once is worth more than memorising it, because the derivation is what shows why standard form is required and why the sign of b has to be carried.
Core ideas
The formula is a theorem with a proof you can carry out yourself, and the proof is “Completing the Square” done with letters.
a, b and c are the coefficients of ax² + bx + c = 0. 3x² = 5x - 2 has b = -5 and c = 2, and neither can be read until the rearrangement is done. Reading them off the equation as printed is the commonest way to get a wrong answer out of a right formula.
If b = -7 then -b = 7 and b² = 49. Both of those catch people, and they catch them in opposite directions: one drops a sign that should survive, the other keeps a sign that should not.
If b² - 4ac is negative the formula asks for the square root of a negative number, and no real number supplies one. That is not a breakdown — it is the formula reporting that there is no real solution, which “The Discriminant” learns to read off before any work is done.
The material
The proof, the one subtlety inside it, and how to use the result without manufacturing sign errors.
“Completing the Square” completes the square on any particular quadratic. Doing it on ax² + bx + c = 0, with letters in place of numbers, does it for all of them at once and leaves a formula behind.
Let a, b and c be real with a ≠ 0. If b² - 4ac ≥ 0, the real solutions of ax² + bx + c = 0 are exactly x = (-b + √(b² - 4ac))/(2a) and x = (-b - √(b² - 4ac))/(2a). If b² - 4ac < 0 there are no real solutions.
Divide through by a, which is allowed because a ≠ 0: x² + (b/a)x + c/a = 0. Move the constant across: x² + (b/a)x = -c/a.
Half the coefficient of x is b/(2a), and its square is b²/(4a²). Add that to both sides: x² + (b/a)x + b²/(4a²) = -c/a + b²/(4a²).
The left side is now (x + b/(2a))². On the right, put both terms over 4a²: since -c/a = -4ac/(4a²), the right side is (b² - 4ac)/(4a²).
4a² is positive, so the right side has the same sign as b² - 4ac. By the square root property there is a real solution exactly when b² - 4ac ≥ 0, and then x + b/(2a) = ±√(b² - 4ac) / √(4a²).
Now √(4a²) = 2|a|, not 2a, because a square root is never negative. But the numerator already carries a ±, so the two numbers produced by dividing by 2|a| are the same two numbers produced by dividing by 2a — listed in the other order when a is negative. Writing 2a is therefore safe, and it is what keeps the formula short.
Subtracting b/(2a) gives x = -b/(2a) ± √(b² - 4ac)/(2a), and the two terms share a denominator: x = (-b ± √(b² - 4ac))/(2a).
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The quantity b² - 4ac appeared without being asked for, and it decides everything about the answer. “The Discriminant” is about that quantity alone.
Three things are worth doing in a fixed order every single time: rearrange into standard form, write a, b and c down with their signs, and compute b² - 4ac on its own line before substituting anything else.
-b was -(-7) = 7, and b² was (-7)² = 49. Writing b = -7 on its own line first is what makes both of those automatic rather than remembered.
This one can be checked independently, because it happens to factor: 2x² - 7x + 3 = (2x - 1)(x - 3), giving x = 1/2 and x = 3. Two methods, computed separately, agreeing.
The formula frequently hands back an unreduced fraction, and the reduction has to divide every term of the numerator, not the convenient one:
√12 = 2√3 by Rational and Radical Expressions, and then the 2 underneath divides into -4 and into 2√3 alike. Cancelling from one term only produces a number that is not a solution: substituting -4 + √3 into x² + 4x + 1 gives 4 - 4√3, which is not 0.
a may be negative and the formula does not mind. -x² + 5x - 6 = 0 has a = -1, b = 5, c = -6, so b² - 4ac = 25 - 4(-1)(-6) = 25 - 24 = 1 and x = (-5 ± 1)/(-2), giving x = 2 and x = 3. Multiplying the equation through by -1 first gives x² - 5x + 6 = 0 and the same two solutions with far less sign handling, which is a free simplification worth taking.
x² - 9 = 0 has b = 0, and the formula gives x = ±√36 / 2 = ±3. Correct, and slower than the square root property, which reads x² = 9 and stops.
The formula always works. It is not always the shortest route, and one line of “The Discriminant” usually tells you which route to take.
Interactive
Substituted, and checked against the other methods
Method
The first two steps hold the errors; the last two are arithmetic.
Every term on one side in descending powers, zero on the other. Nothing can be read off the equation until this is done.
3x² - 5x - 2 = 0 gives a = 3, b = -5, c = -2. That line costs three seconds and removes the two commonest errors in the course.
(-5)² - 4(3)(-2) = 25 + 24 = 49. Brackets round a negative coefficient cost nothing. If the result is negative, stop: there is no real solution.
√49 = 7, so x = (5 ± 7)/6, which is 2 or -1/3. When the radical does not reduce to an integer, cancel only by a factor that divides the whole numerator.
One that also factors, one that does not, and one with a negative leading coefficient.
In C the term -4ac was -4(-1)(-6), a product of three negative factors and therefore negative: -24. That single sign is the difference between b² - 4ac = 1 and b² - 4ac = 49, and between the answers 2 and 3 and the answers -1 and 6.
B is the case the formula exists for. x² + 4x + 1 has no integer factorisation, the answers are irrational, and completing the square would have produced them in about the same number of lines — which is precisely what the derivation predicts, since the formula is that completion, already done.
For a faded formula pass, solve 3x² + 2x - 7 = 0. The supplied coefficient line is a = 3, b = 2, c = -7. Compute D with the sign of c visible, substitute, simplify the radical and divide the whole numerator. Check against x = (-1 ± √22)/3; a denominator of 6 left in the final form means the common factor in the numerator was not removed.
Take ax² + bx + c = 0 and complete the square on it from memory, reaching the formula in six or seven lines. Then use it with a, b and c written out separately every time, and reduce the result by dividing the whole numerator.
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