Two methods that work when factoring will not — and one is just the other, done once and for all.
Complete the square
What number must be added to x² + 12x to make it a perfect square trinomial?
Halve the coefficient of x and square it: half of 12 is 6, and 6² = 36. That gives x² + 12x + 36 = (x + 6)². A stops at the halving and forgets to square. D squares without halving. Both steps, in that order, every time.
x² + bx needs (b2)² to become a perfect square.
The name is literal: you are completing a square that is missing its corner. Halving b splits the rectangle into two equal strips; squaring gives the corner they leave behind.
x² + 6x − 7 = 0
move the constant
x² + 6x = 7
half of 6 is 3, 3² = 9
x² + 6x + 9 = 16
add 9 to BOTH sides
(x + 3)² = 16
now it is a square
x + 3 = ±4, so x = 1 or −7
extract roots
Adding 9 to the left only would change the equation. And if the leading coefficient is not 1, divide the whole equation by it before halving anything.
x = −b ± √b² − 4ac2a
3x² + 5x − 2 = 0
a = 3, b = 5, c = −2
b² − 4ac = 25 + 24 = 49
the discriminant; mind the double negative
x = −5 ± 76
√49 = 7
x = 13 or x = −2
two branches
Here c is −2, so −4ac becomes −4(3)(−2) = +24. Dropping that minus gives 1 instead of 49 and ruins everything after it. The part under the root, b² − 4ac, has its own name: the discriminant.
x² − 4x − 1 = 0
will not factor
b² − 4ac = 16 + 4 = 20
not a perfect square
x = 4 ± √202
√20 = 2√5
x = 2 ± √5
≈ 4.24 or −0.24
A discriminant that is not a perfect square is the sign the equation was never going to factor. That is when the formula earns its keep.
The formula comes from completing the square on ax² + bx + c = 0 in general. Learning both is not duplication — completing the square is the only one that also hands you the vertex.
With no other side to balance, add the square number and subtract it straight back.
Write x² − 10x + 18 in the form (x + h)² + k.
x² − 10x + 25 − 25 + 18
half of −10 is −5; (−5)² = 25
(x − 5)² − 25 + 18
fold the first three terms
(x − 5)² − 7
check: x² − 10x + 25 − 7 = x² − 10x + 18 ✓
Find the vertex of y = x² + 8x + 11.
y = (x + 4)² − 16 + 11
half of 8 is 4; take 16 back off
y = (x + 4)² − 5
smallest when the bracket is 0
vertex (−4, −5)
check: 16 − 32 + 11 = −5 ✓
The bracket (x + 4) is zero when x = −4, so that is where the vertex sits. Copying the +4 straight out of the bracket puts the vertex on the wrong side of the y-axis.
The same moves, in the three places they look different.
An odd b: x² + 3x − 1 = 0
x² + 3x + 94 = 1 + 94
half of 3 is 32; (32)² = 94
(x + 32)² = 134
44 + 94
x = −3 ± √132
√4 = 2 on the bottom
Keep the half as a fraction; rounding 32 to 2 ruins the square.
A leading 3: 3x² − 12x + 6 = 0
x² − 4x + 2 = 0
divide every term by 3
(x − 2)² = 2
x² − 4x + 4 = −2 + 4
x = 2 ± √2
formula check: 12 ± √726 = 2 ± √2 ✓
Divide by a first, so x² stands alone; only then halve the new b, −4.
A negative discriminant: x² + 2x + 5 = 0
b² − 4ac = 4 − 20 = −16
a = 1, b = 2, c = 5
(x + 1)² = −4
the same fact, by completing the square
no real roots
a square cannot be −4
Stop at the discriminant: a negative one means there is nothing to take the root of.
Expect three kinds of item: the number that completes a square, a quadratic solved by the formula with a root to simplify, and an expression rewritten as (x + h)² + k to read off its vertex or minimum. The fastest route is to write a, b and c with their signs, find the discriminant first, and simplify the root before dividing. The 10-second check: expand your (x + h)² + k back to the original, or add your two roots and compare with −ba.
Use the formula
Solve x² − 6x + 4 = 0.
a = 1, b = −6, c = 4. The discriminant is 36 − 16 = 20, so x = 6 ± √202 = 6 ± 2√52 = 3 ± √5. B forgets to divide by 2a. D has −b wrong — b is −6, so −b is +6. Check: 3 + √5 ≈ 5.24, and 5.24² − 6(5.24) + 4 ≈ 0 ✓
Hint: b is negative, so −b is positive. Simplify √20 to 2√5 before dividing, and the 2s cancel neatly.
