Deriving the quadratic formula

Start from the general equation, with a \neq 0:

Move the constant. Completing the square works on the x terms alone, so subtract c from both sides:

Divide by a. Completing the square needs the coefficient of x^2 to be 1. Dividing both sides by a is allowed precisely because a \neq 0, and it keeps the same solutions, since multiplying by a undoes it:

Add half the coefficient of x, squared, to both sides. Half of \displaystyle \frac{b}{a} is \displaystyle \frac{b}{2a}, and its square is \displaystyle \frac{b^2}{4a^2}:

Write the left side as a square. The number inside the brackets is the half, \displaystyle \frac{b}{2a}. On the right, put both fractions over 4a^2 by writing \displaystyle \frac{c}{a} = \frac{4ac}{4a^2}:

This line is the heart of the derivation. The left side is a square, so it is never negative when x is real. The right side has a positive denominator, 4a^2, so its sign is the sign of b^2 - 4ac. That is why the discriminant decides whether real solutions exist.

Take square roots, keeping both signs. By the square root property,

Now \displaystyle \sqrt{4a^2} = 2|a|, which is 2a when a is positive and -2a when a is negative. It does not matter which: dividing by -2a instead of 2a only swaps which solution goes with + and which with -, and the \pm lists both anyway. So

Subtract \displaystyle \frac{b}{2a} from both sides. The two fractions already share the denominator 2a:

Every step can be run backwards: add c back, multiply by a, subtract what was added, square. So no step created a false solution or lost a true one, and the formula gives exactly the solutions of the original equation.

The completed-square result is the same work done on the expression instead of the equation. An expression has no second side, so instead of dividing by a, factor it out of the x terms, then add and subtract \displaystyle \frac{b^2}{4a^2} inside the brackets, as in completing the square with a leading coefficient. Multiply out the right side to check it.

For the sum and product, add the two solutions: the square roots cancel, leaving \displaystyle \frac{-2b}{2a} = -\frac{b}{a}. Multiply them, using (p + q)(p - q) = p^2 - q^2:

1. \displaystyle x = -3 \pm \sqrt{5}. Move the 4 and add 9: (x + 3)^2 = 5. The formula: \displaystyle \frac{-6 \pm \sqrt{36 - 16}}{2} = \frac{-6 \pm 2\sqrt{5}}{2}, the same.
2. \displaystyle x = \frac{2 \pm \sqrt{10}}{2}. Divide by 2: \displaystyle x^2 - 2x = \frac{3}{2}; add 1: \displaystyle (x - 1)^2 = \frac{5}{2}, and \displaystyle \sqrt{\frac{5}{2}} = \frac{\sqrt{10}}{2}. The formula: \displaystyle \frac{4 \pm \sqrt{40}}{4} = \frac{4 \pm 2\sqrt{10}}{4}, the same.
3. \displaystyle x = \frac{1}{3} or x = -1. Divide by 3: \displaystyle x^2 + \frac{2}{3}x = \frac{1}{3}; add \displaystyle \frac{1}{9}: \displaystyle \left(x + \frac{1}{3}\right)^2 = \frac{4}{9}. The formula: \displaystyle \frac{-2 \pm \sqrt{16}}{6} = \frac{-2 \pm 4}{6}.
4. \displaystyle x = \frac{1}{2} or x = -3. Here \displaystyle \frac{b}{2a} = \frac{5}{4} and \displaystyle \frac{b^2 - 4ac}{4a^2} = \frac{25 + 24}{16} = \frac{49}{16}, so \displaystyle \left(x + \frac{5}{4}\right)^2 = \frac{49}{16} and \displaystyle x + \frac{5}{4} = \pm\frac{7}{4}.
5. \displaystyle x = \frac{1}{2} \pm \sqrt{2}. Here \displaystyle \frac{b}{2a} = -\frac{1}{2} and \displaystyle \frac{b^2 - 4ac}{4a^2} = \frac{16 + 112}{64} = 2, so \displaystyle \left(x - \frac{1}{2}\right)^2 = 2.
6. \displaystyle x = \frac{-2 \pm \sqrt{10}}{3}. Multiply by 12: 36x^2 + 48x - 24 = 0. Move the 24 and add b^2 = 16: (6x + 4)^2 = 40, so \displaystyle 6x + 4 = \pm 2\sqrt{10}.
7. x = 1 or \displaystyle x = -\frac{2}{5}. Multiply by 20: 100x^2 - 60x - 40 = 0. Move the 40 and add b^2 = 9: (10x - 3)^2 = 49, so 10x - 3 = \pm 7.
8. \displaystyle x = \frac{-b \pm \sqrt{b^2 - 4c}}{2}. Move c and add \displaystyle \frac{b^2}{4}: \displaystyle \left(x + \frac{b}{2}\right)^2 = \frac{b^2}{4} - c = \frac{b^2 - 4c}{4}, so \displaystyle x + \frac{b}{2} = \pm\frac{\sqrt{b^2 - 4c}}{2}. With a = 1 the general formula reads the same.
9. \displaystyle x = \frac{-k \pm \sqrt{k^2 - ac}}{a}, and for 3x^2 + 8x + 2 = 0 (so k = 4), \displaystyle x = \frac{-4 \pm \sqrt{10}}{3}. Divide by a: \displaystyle x^2 + \frac{2k}{a}x = -\frac{c}{a}. Half the coefficient of x is \displaystyle \frac{k}{a}, so \displaystyle \left(x + \frac{k}{a}\right)^2 = \frac{k^2 - ac}{a^2}. For the example, k^2 - ac = 16 - 6 = 10.
10. x = 1 or \displaystyle x = -\frac{5}{2}, as in Example 1. The top becomes (-b)^2 - \left(b^2 - 4ac\right) = 4ac, by (p + q)(p - q) = p^2 - q^2, so \displaystyle x = \frac{4ac}{2a\left(-b \mp \sqrt{b^2 - 4ac}\right)} = \frac{2c}{-b \mp \sqrt{b^2 - 4ac}}. The new bottom is 0 only if b^2 = b^2 - 4ac, that is only if c = 0. For the check, 2c = -10 and \displaystyle \sqrt{49} = 7: \displaystyle \frac{-10}{-3 - 7} = 1 and \displaystyle \frac{-10}{-3 + 7} = -\frac{5}{2}.
11. \displaystyle x = \frac{p \pm \sqrt{p^2 + 4q}}{2}. Subtract px and add \displaystyle \frac{p^2}{4}: \displaystyle \left(x - \frac{p}{2}\right)^2 = \frac{p^2 + 4q}{4}. Since q > 0, \displaystyle \sqrt{p^2 + 4q} > \sqrt{p^2} = p, so the root with + is positive and the root with - is negative. Al-Khwarizmi’s rule gave only the positive one.
12. \displaystyle \left(x - \frac{5}{6}\right)^2 = -\frac{23}{36}, so there is no real solution. Here \displaystyle \frac{b}{2a} = -\frac{5}{6} and \displaystyle \frac{b^2 - 4ac}{4a^2} = \frac{25 - 48}{36}. The left side is a square, never negative for real x, and the right side is negative.
13. Sum \displaystyle \frac{9}{2}, product 2. These are \displaystyle -\frac{b}{a} and \displaystyle \frac{c}{a}. Solving: b^2 - 4ac = 81 - 32 = 49, so \displaystyle x = \frac{9 \pm 7}{4}, which is 4 or \displaystyle \frac{1}{2}; and \displaystyle 4 + \frac{1}{2} = \frac{9}{2}, \displaystyle 4 \cdot \frac{1}{2} = 2.
14. b = -6 and c = 7. The center is \displaystyle -\frac{b}{2} = 3, so b = -6. The distance is \displaystyle \frac{\sqrt{b^2 - 4c}}{2} = \sqrt{2}, so b^2 - 4c = 8, that is 36 - 4c = 8. Check: x^2 - 6x + 7 = 0 has solutions \displaystyle \frac{6 \pm \sqrt{8}}{2} = 3 \pm \sqrt{2}.
15. 2(x + 3)^2 - 11. Here \displaystyle \frac{b}{2a} = 3 and \displaystyle \frac{b^2 - 4ac}{4a} = \frac{144 - 56}{8} = 11.
16. x = m + 1 or x = m - 1. With a = 1, b = -2m, c = m^2 - 1 the discriminant is 4m^2 - 4m^2 + 4 = 4, so \displaystyle x = \frac{2m \pm 2}{2}. It does not depend on m because the equation is already (x - m)^2 = 1: changing m slides both solutions along together without changing the distance between them.
17. \displaystyle x = 3 \pm \sqrt{y + 4}. Write x^2 - 6x + (5 - y) = 0; then b^2 - 4ac = 36 - 20 + 4y = 4(y + 4), so \displaystyle x = \frac{6 \pm 2\sqrt{y + 4}}{2}.
18. \displaystyle r = -h + \sqrt{h^2 + A/\pi}. As a quadratic in r: \pi r^2 + 2\pi h r - A = 0, so \displaystyle r = \frac{-2\pi h \pm \sqrt{4\pi^2 h^2 + 4\pi A}}{2\pi}, which simplifies to \displaystyle -h \pm \sqrt{h^2 + A/\pi}. The root with - is negative, and a radius cannot be.
19. Dividing by 0 is not allowed, and the formula would have 0 underneath. With a = 0 the equation is bx + c = 0, a linear equation, solved as \displaystyle x = -\frac{c}{b} when b \neq 0.
20. 2|a| and 2a differ only in sign. So using -2a in place of 2a gives \displaystyle \pm\frac{\sqrt{b^2 - 4ac}}{-2a}, which is the same two numbers as \displaystyle \pm\frac{\sqrt{b^2 - 4ac}}{2a}, with + and - swapped. The \pm lists both.
21. Write \displaystyle s = \sqrt{b^2 - 4ac}, so s^2 = b^2 - 4ac. Then \displaystyle ax^2 = \frac{b^2 - 2bs + s^2}{4a}, \displaystyle bx = \frac{-2b^2 + 2bs}{4a} and \displaystyle c = \frac{4ac}{4a}. Adding over 4a, the bs terms cancel and the top is s^2 - b^2 + 4ac = 0. The same working with -s in place of s checks the other solution.
22. The average is \displaystyle \frac{1}{2}(x_1 + x_2) = \frac{1}{2}\left(-\frac{b}{a}\right) = -\frac{b}{2a}, because the two square roots cancel when the solutions are added. On the graph it is where the axis of symmetry crosses the x-axis: the two x-intercepts are mirror images across it.
23. If b^2 - 4ac = m^2 for a whole number m \ge 0, then \displaystyle \sqrt{b^2 - 4ac} = m, and the formula gives \displaystyle \frac{-b + m}{2a} and \displaystyle \frac{-b - m}{2a}. Each is a whole number over the whole number 2a, which is not 0, so a fraction.