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Completing the Square & the Quadratic Formula

When factoring fails you need a method that always works. Completing the square forces any quadratic into a perfect square, and running that process once on the general form produces the quadratic formula — the master key.

Completing the square

[[completing-the-square|Completing the square]] turns any quadratic into a perfect-square trinomial plus a leftover number, so you can finish with the square-root property. The trick rests on the pattern (x + p)^2 = x^2 + 2px + p^2. To rebuild that pattern, take half of the x-coefficient and square it — that is the number you add (and subtract) to keep the equation balanced.

Solve  x^2 + 6x - 7 = 0  by completing the square.

    x^2 + 6x = 7              (move constant right)
  half of 6 is 3, and 3^2 = 9:
    x^2 + 6x + 9 = 7 + 9       (add 9 to BOTH sides)
        (x + 3)^2 = 16          (left side is a perfect square)
         x + 3 = ±4             (square-root property)
         x = -3 ± 4
         x = 1   or   x = -7
Half the middle coefficient, squared, completes the square.

The quadratic formula: completing the square, once and for all

Completing the square works on every quadratic — so what if you do it once on the general form a x^2 + b x + c = 0, with letters instead of numbers? The answer that falls out is the [[quadratic-formula|quadratic formula]], a single expression that solves any quadratic the moment you read off a, b, c. It is not a separate trick; it is completing the square, frozen into a formula.

Quadratic formula:

              -b ± sqrt(b^2 - 4ac)
        x  =  --------------------
                      2a

Solve  2x^2 + 3x - 5 = 0   (a = 2, b = 3, c = -5):

    x = ( -3 ± sqrt(3^2 - 4·2·(-5)) ) / (2·2)
    x = ( -3 ± sqrt(9 + 40) ) / 4
    x = ( -3 ± sqrt(49) ) / 4
    x = ( -3 ± 7 ) / 4
    x = 4/4 = 1     or     x = -10/4 = -5/2
Read off a, b, c and substitute — the formula never fails.