Nib asked
Why does the quadratic formula always work?
The quadratic formula is just completing the square done once, for every quadratic at the same time; it gives the roots of any equation with a not zero, though they are real only when b squared minus 4ac is not negative.
6 steps, drawn by Nib. Read it here, or watch him draw it. Updated .
1a missing corner
The quadratic formula is not a trick pulled out of a hat. It is one technique, completing the square, carried out once on the general equation. To see the technique, take x² + 6x = 7. Draw x² as a square with side x, and 6x as two strips of width 3 along two sides. Together they make an L-shape with a corner missing.
x² plus 6x is a square with a corner missing.
2fill the corner
Add a 3 by 3 corner, area 9, to complete the big square, and add 9 to the other side too. Now (x + 3)² = 7 + 9 = 16, so x + 3 = 4 or x + 3 = −4, giving x = 1 or x = −7. Check: 1 + 6 = 7, and 49 − 42 = 7.
Adding the missing corner turns the left side into one perfect square.
Check x = −7 in x² + 6x = 7.
Show the answer
49 − 42 = 7, so it works.
3any quadratic
Take ax² + bx + c = 0 with a not zero. Divide by a: x² + (b/a)x = −c/a. The missing corner is (b/2a)², so add it to both sides. The left becomes (x + b/2a)² and the right becomes b²/4a² − c/a = (b² − 4ac) / 4a².
Divide by a, add (b/2a)² to both sides: a perfect square appears.
Why must a not be zero?
Show the answer
Dividing by a is the first step, and you can't divide by zero; with a = 0 it isn't a quadratic.
4the formula
Take square roots of both sides, keeping both signs: x + b/2a = ± √(b² − 4ac) / 2a. Subtract b/2a: x = (−b ± √(b² − 4ac)) / 2a. Each step is reversible algebra, so any solution of the original equation shows up in the formula and the formula's answers always solve it.
x = (−b ± √(b² − 4ac)) / 2a, for every quadratic with a not zero.
5under the root
The expression b² − 4ac under the root is called the discriminant. If it is positive, the formula gives two different real solutions, and the parabola y = ax² + bx + c crosses the x-axis twice. If it is zero, the two solutions coincide and the parabola just touches the axis. If it is negative, there is no real square root, so the parabola never reaches the axis and the solutions are complex numbers instead.
The sign of b² − 4ac says how many times the parabola meets the axis.
a common mix-up'Always works' needs two small print lines: a must not be zero, and a negative b² − 4ac gives no real answers (complex ones).
What does a negative b² − 4ac tell you about the parabola?
Show the answer
It never crosses the x-axis, so the equation has no real solutions.
6the keepsake
So yes, it always works, for any quadratic with a not zero. Real answers when that discriminant isn't negative, complex ones when it is. Same square, completed once.
The formula is one completed square, worked out once so you never have to do it again.