Quadratic Formula¶

The Quadratic Formula is a formula of which can be used to solve a quadratic. It can be derived from $ax^2 + bx + c = 0$ , using a method involving completing the square.

Formula¶

The formula is the following:

$x = \frac{-b \pm\sqrt{b^2 - 4ac}}{2a}$

Derivation¶

As previously noted, the quadratic formula can be derived from $ax^2 + bx + c = 0$ using a method involving completing the square.

To prepare the equation for completing the square, you'll need to go about making the equation start with $x^2$ - simply by dividing the equation through by $a$ . Then you can go about giving the equation a value, other than 0, to be equal to - use $\frac{c}{a}$ !

$\begin{align} ax^2 + bx + c & = 0 \\ x^2 + \frac{b}{a}x + \frac{c}{a} & = 0 \\ x^2 + \frac{b}{a}x & = - \frac{c}{a} \end{align}$

The equation is now in a position to be put into a completed square form. Expanding out $b$ from the completed square form - $(\frac{b}{2a})^2$ in this case.

$\begin{align} x^2 + \frac{b}{a}x & = - \frac{c}{a} \\ (x + \frac{b}{2a})^2 - (\frac{b}{2a})^2 & = - \frac{c}{a} \\ (x + \frac{b}{2a})^2 - \frac{b^2}{4a^2} & = - \frac{c}{a} \end{align}$

The equation should now be in a position to be made equal to $(x + \frac{b}{2a})^2$ in preparation of solving the quadratic. Having done so, a common denominator ( $4a^2$ ) should be found of the fractions to merge them.

$\begin{align} (x + \frac{b}{2a})^2 - \frac{b^2}{4a^2} & = - \frac{c}{a} \\ (x + \frac{b}{2a})^2 & = \frac{b^2}{4a^2} - \frac{c}{a} \\ (x + \frac{b}{2a})^2 & = \frac{b^2}{4a^2} - \frac{4ac}{4a^2} \\ (x + \frac{b}{2a})^2 & = \frac{b^2 - 4ac}{4a^2} \end{align}$

The equation is now ready to be solved, by square rooting, and subtracting $\frac{b}{2a}$ .

$\begin{align} (x + \frac{b}{2a})^2 & = \frac{b^2 - 4ac}{4a^2} \\ x + \frac{b}{2a} & = \frac{\pm\sqrt{b^2 - 4ac}}{2a} \\ x & = \frac{\pm\sqrt{b^2 - 4ac}}{2a} - \frac{b}{2a} \\ x & = \frac{-b \pm\sqrt{b^2 - 4ac}}{2a} \end{align}$

Alternate Form¶

As with all formulas there is potential for other forms of the same thing.

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

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