Build a perfect square on purpose, then square-root both sides.
Solve for x
x² + 6x − 7 = 0
Half of 6 is 3 and 3² = 9, so x² + 6x + 9 = 7 + 9 makes (x + 3)² = 16. Then x + 3 = ±4, giving x = 1 or x = −7. D stops at ±4 without moving the 3 back across.
Keep the x terms together on the left.
x² + 6x − 7 = 0
the question
x² + 6x = 7
the left is now incomplete
x² + 6x is not yet a perfect square. The plan is to add exactly the right number to make it one — and to add it to the other side too.
That single instruction is the whole method.
b
6
b ÷ 2
3
add this
9
x² + 6x + 9 = 7 + 9
both sides, always
x² + 6x + 9 = 16
the right tidies up
Adding 9 to the left only changes the equation into a different one. Whatever you build on one side must be paid for on the other.
The number inside is the half you just took — the 3, not the 9.
x² + 6x + 9
(x + 3)2 = 16
The 9 is literally the missing corner — which is where “completing the square” gets its name.
(x + 3)² = 16
from Step 3
x + 3 = ± 4
both 4² and (−4)² give 16
x + 3 = 4
x = 1
x + 3 = −4
x = −7
Taking only +4 loses a whole root. Every square root of a number has two signs.
The halve-and-square trick only works with a bare x².
2x² + 8x = 10
a is 2
x² + 4x = 5
÷ 2 across the whole line
x² + 4x + 4 = 9
half of 4 is 2, squared is 4
(x + 2)² = 9
x = 1 or x = −5
Dividing by a often leaves a fraction on the right. 2x² − 4x − 1 = 0: x² − 2x = 12, then x² − 2x + 1 = 12 + 1, so (x − 1)² = 32. Root top and bottom, then clear the root from the bottom: x − 1 = ±√3√2 = ±√62, so x = 1 ± √62.
Run this method on ax² + bx + c = 0 with letters instead of numbers and the quadratic formula falls out. That is where the formula comes from.
There is no other side to balance, so we add the square and subtract it straight back.
x² − 8x + 21
half of −8 is −4; (−4)² = 16
x² − 8x + 16 − 16 + 21
+16 and −16 change nothing
(x − 4)² + 5
fold the first three terms
least value 5, at x = 4
a square is never below 0
This is the vertex form of y = x² − 8x + 21: the vertex is (4, 5), the minimum value is 5, and the range is y ≥ 5.
The same four moves; these are the three places students stall.
An odd b, so fractions appear: x² + 5x + 1 = 0
x² + 5x = −1
move the constant
x² + 5x + 254 = −1 + 254
half of 5 is 52; (52)² = 254
(x + 52)² = 214
−44 + 254 = 214
x + 52 = ±√212
√4 = 2 on the bottom
x = −5 ± √212
one fraction over 2
Keep halves as fractions, never as rounded decimals. Writing −1 as −44 makes the right side one easy sum.
Rounding 52 down to 2 and adding 4 leaves x² + 5x + 4, which is not a perfect square. With an odd b the number to add is always a fraction over 4.
A negative x² when solving: −x² + 4x + 2 = 0
x² − 4x − 2 = 0
divide every term by −1
x² − 4x + 4 = 2 + 4
half of −4 is −2; (−2)² = 4
(x − 2)² = 6
fold
x = 2 ± √6
6 is not a perfect square
In an equation we may divide the whole line by a = −1. Only in an expression, as in the next step, must the −1 be factored out instead.
The right side comes out negative: x² + 2x + 6 = 0
x² + 2x + 1 = −6 + 1
move 6, then add 1² = 1
(x + 1)² = −5
a square equal to a negative
no real roots
no real number squares to −5
The method tells us itself when there is no answer: stop at the negative, do not take its square root.
The constant stays outside; complete the square inside the bracket.
3x² + 12x + 5 = 3(x² + 4x) + 5
factor 3 from the x terms
= 3(x² + 4x + 4) − 12 + 5
the 4 inside is worth 3 × 4 = 12
= 3(x + 2)² − 7
minimum −7, at x = −2
−x² + 6x − 4 = −(x² − 6x) − 4
factor out −1
= −(x² − 6x + 9) + 9 − 4
the 9 inside is worth −9, so add 9 back
= −(x − 3)² + 5
maximum 5, at x = 3
Inside 3( ), the added 4 is really 12. Writing 3(x + 2)² + 5 − 4 = 3(x + 2)² + 1 forgets the 3 — the most common wrong k.
Exam items use completing the square for its form more than for solving: write a(x − h)² + k, read off a minimum, a maximum or a range, or match coefficients. The 10-second check is to expand your answer in your head — a(x − h)² + k must give back the original b and c.
Solve by completing the square
x² − 10x + 9 = 0
Half of −10 is −5 and (−5)² = 25, so (x − 5)² = −9 + 25 = 16. Then x − 5 = ±4, giving x = 9 or 1. A forgets to add the 5 back — the ± step gives you x − 5, not x.
Hint: b is −10, so half of it is −5 and the bracket will read (x − 5). Squaring −5 still gives +25, which goes on both sides.
