Every way to solve a quadratic
Four methods, one decision rule for picking between them.
Try factoring first — it's fastest when the numbers are nice. If it
doesn't factor cleanly but the equation is already (something)^2 = k,
the square root property is instant. Completing the square always works
but touches the most steps. The quadratic formula is the guaranteed
fallback — never wrong, so reach for it the moment the others feel slow.
If A and B are real numbers (or real expressions) and A \cdot B = 0, then A = 0 or B = 0 (or both).
Requires the equation to already equal zero and factor cleanly.
Show proof
Suppose A \cdot B = 0 and A \neq 0. Since A \neq 0, it has a multiplicative inverse \frac{1}{A}. Multiply both sides by \frac{1}{A}: \frac{1}{A}(A \cdot B) = \frac{1}{A} \cdot 0 The left side simplifies to B (since \frac{1}{A} \cdot A = 1) and the right side is 0, so B = 0. By the same argument with the roles swapped, if B \neq 0 then A = 0. So at least one of the two factors must be 0.
If u^2 = k with k \ge 0, then u = \sqrt{k} or u = -\sqrt{k}, abbreviated u = \pm\sqrt{k}. If k < 0, there are no real solutions.
Fastest when the equation is already isolated as (something)² = k.
To solve x^2 + bx + c = 0: move the constant right, then add \left(\frac{b}{2}\right)^2 to both sides. The left side becomes the perfect square \left(x + \frac{b}{2}\right)^2, and the square root property finishes the job. (If a \ne 1, divide the equation by a first.)
Always works — rebuilds any quadratic into "something squared" form.
For a \neq 0, x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.
The universal fallback — solves every quadratic in one shot.
Show proof
Start from the general quadratic ax^2 + bx + c = 0 and divide every term by a (legal since a \neq 0): x^2 + \frac{b}{a}x + \frac{c}{a} = 0
Move the constant term to the right side: x^2 + \frac{b}{a}x = -\frac{c}{a}
Complete the square: take half of the x-coefficient, \frac{b}{2a}, square it, and add it to both sides. x^2 + \frac{b}{a}x + \left(\frac{b}{2a}\right)^2 = \left(\frac{b}{2a}\right)^2 - \frac{c}{a}
The left side is now a perfect-square trinomial: \left(x + \frac{b}{2a}\right)^2 = \frac{b^2}{4a^2} - \frac{c}{a} = \frac{b^2 - 4ac}{4a^2}
Take the square root of both sides — remembering a square root has two signs (strictly, \sqrt{4a^2} = 2|a| since a could itself be negative, but the very next \pm absorbs that sign either way, so writing 2a in the denominator costs nothing): x + \frac{b}{2a} = \pm\frac{\sqrt{b^2 - 4ac}}{2a}
Subtract \frac{b}{2a} from both sides: x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
For ax^2 + bx + c = 0 with a \neq 0, let D = b^2 - 4ac. Then:
- D > 0: two distinct real solutions (rational if D is a perfect square, irrational otherwise).
- D = 0: exactly one repeated real solution, x = -\frac{b}{2a}.
- D < 0: no real solutions; instead, two complex conjugate solutions.
Predicts how many real solutions to expect before you solve.
Show proof
The Quadratic Formula gives every solution of ax^2+bx+c=0 as x = \frac{-b \pm \sqrt{D}}{2a}, \qquad D = b^2 - 4ac, and the three cases fall directly out of what \sqrt{D} does under the \pm.
D > 0. \sqrt{D} is a positive real number, and +\sqrt D \ne -\sqrt D, so the formula produces two distinct real values. If D is a perfect square, \sqrt D is a whole number and the whole expression is a ratio of integers — rational. If D is not a perfect square, \sqrt D is irrational, and an irrational number added to the rational -b stays irrational, so both solutions are irrational.
D = 0. The \pm\sqrt D term vanishes entirely, collapsing the two branches into the single value x = -\frac{b}{2a} — one repeated solution.
D < 0. No real number squares to a negative, so \sqrt D has no real value and the formula names no real solution. Writing \sqrt D = i\sqrt{-D} instead produces the two complex values x = \frac{-b \pm i\sqrt{-D}}{2a}, which are conjugates of each other since they differ only in the sign in front of the imaginary part.
For f(x) = ax^2 + bx + c with a \neq 0, the vertex occurs at x = -\frac{b}{2a}, and f\!\left(-\frac{b}{2a}\right) is the minimum value of f when a > 0 and the maximum value when a < 0.
Not a solving method, but the same completed-square work locates the vertex.