# derivation of quadratic formula

Since $A$ is nonzero, we can divide by $A$ and obtain the equation

$${x}^{2}+bx+c=0,$$ |

where $b=\frac{B}{A}$ and $c=\frac{C}{A}$. This equation can be written as

$${x}^{2}+bx+\frac{{b}^{2}}{4}-\frac{{b}^{2}}{4}+c=0,$$ |

so completing the square, i.e., applying the identity ${(p+q)}^{2}={p}^{2}+2pq+{q}^{2}$, yields

$${\left(x+\frac{b}{2}\right)}^{2}=\frac{{b}^{2}}{4}-c.$$ |

Then, taking the square root of both sides, and solving for $x$, we obtain
the solution formula^{}

$x$ | $=$ | $-{\displaystyle \frac{b}{2}}\pm \sqrt{{\displaystyle \frac{{b}^{2}}{4}}-c}$ | ||

$=$ | $\frac{B}{2A}}\pm \sqrt{{\displaystyle \frac{{B}^{2}}{4{A}^{2}}}-{\displaystyle \frac{C}{A}}$ | |||

$=$ | $\frac{-B\pm \sqrt{{B}^{2}-4AC}}{2A}},$ |

and the derivation is completed.

A slightly less intuitive but more aesthetically pleasing approach to this derivation can be achieved by multiplying both sides of the equation

$a{x}^{2}+bx+c=0$ |

by $4a$, resulting in the equation

$4{a}^{2}{x}^{2}+4abx+{b}^{2}={b}^{2}-4ac,$ |

in which the left-hand side can be expressed as ${(2ax+b)}^{2}$. From here, the proof is identical.

Title | derivation of quadratic formula |
---|---|

Canonical name | DerivationOfQuadraticFormula |

Date of creation | 2013-03-22 11:56:44 |

Last modified on | 2013-03-22 11:56:44 |

Owner | mathcam (2727) |

Last modified by | mathcam (2727) |

Numerical id | 12 |

Author | mathcam (2727) |

Entry type | Proof |

Classification | msc 12D10 |

Related topic | QuadraticFormula |

Related topic | QuadraticEquationInMathbbC |