FormulaProved

Quadratic formula

Statement

The solutions of $ax^2+bx+c=0$ with $a\neq 0$ are $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ .

Why is it true?

Completing the square turns any quadratic into a perfect square plus a leftover constant, so solving it is the same as solving $(x+\text{something})^2 = \text{number}$ . The formula just records that leftover constant — the discriminant $b^2-4ac$ — and its two square roots give the two places where the parabola crosses the x-axis (or shows there are none, in the real numbers, when the discriminant is negative).

Proof sketch

Divide by $a$ and complete the square: $x^2+\tfrac{b}{a}x+\tfrac{c}{a}=\left(x+\tfrac{b}{2a}\right)^2-\tfrac{b^2-4ac}{4a^2}=0$ . Isolating the squared term and taking square roots of both sides gives $x+\tfrac{b}{2a}=\pm\tfrac{\sqrt{b^2-4ac}}{2a}$ , hence the formula.

Stated by

stated 628Brahmagupta stated 628Muhammad ibn Musa al-Khwarizmi

Topics that use this theorem

Quadratic equations

Related theorems

Vieta's formulas Fundamental theorem of algebra

Step-by-step proofs

No step-by-step proof yet for this theorem.

References

David M. Burton (2011). The History of Mathematics: An Introduction