Completing the Square: the Constant Needed

Halve the coefficient of x, square it, and add: that single instruction turns x² + bx into the perfect square (x + b/2)². Worked example: x² + 10x needs k = (10/2)² = 25, giving (x + 5)². Once an expression is a perfect square you can take a square root of both sides and finish the problem in two lines, which is exactly how the quadratic formula itself is derived and how vertex form a(x − h)² + k is reached.

The name is literal. A Babylonian scribe on tablet BM 13901, around 1800 BCE, solved "I added the area and the side of my square and got 45 (in base 60, i.e. 3/4)" by halving the side coefficient, squaring, and adjusting — arriving at ½ with no algebraic symbols at all. Al-Khwārizmī in ninth-century Baghdad drew the same argument as an actual diagram: a square of side x with two rectangles of width b/2 glued to adjacent sides, leaving a missing corner of area (b/2)² to be filled in. Two traps. First, the leading coefficient must be 1 before you start — for 3x² + 12x, factor out the 3 to get 3(x² + 4x) and complete the square inside. Second, adding k changes the expression, so in an equation you must add it to both sides, and in an expression you must immediately subtract it again to keep the value unchanged.