Quadratic Formula

The quadratic formula, first discovered by the Babylonians four thousand years ago, gives you a surefire way to solve quadratic equations of the form

$0 = a x^{2} + b x + c$ .

Plugging in the values of $a, b, and c$ , you will get the desired values of $x$ .

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

If the expression under the square root sign ( $b^{2} - 4 a c$ , also called the discriminant) is negative, then there are no real solutions. (You need complex numbers to deal with this case properly. These are usually taught in Algebra $2$ .)

If the discriminant is zero, there is only one solution. If the discriminant is positive, then the $\pm$ symbol means you get two answers.

Example 1:

Solve the quadratic equation.

$x^{2} - x - 12 = 0$

Here $a = 1, b = - 1, and c = - 12$ . Substituting, we get:

$x = \frac{- (- 1) \pm \sqrt{{(- 1)}^{2} - 4 (1) (- 12)}}{2 (1)}$

Simplify.

$x = \frac{1 \pm \sqrt{49}}{2}$

The discriminant is positive, so we have two solutions:

$x = \frac{8}{2}$ and $- \frac{6}{2}$

$x = 4$ and $x = - 3$

In this example, the discriminant was $49$ , a perfect square, so we ended up with rational answers. Often, when using the quadratic formula, you end up with answers which still contain radicals.

Example 2:

Solve the quadratic equation.

$3 x^{2} + 2 x + 1 = 0$

Here $a = 3, b = 2, and c = 1$ . Substituting, we get:

$x = \frac{- 2 \pm \sqrt{2^{2} - 4 (3) (1)}}{2 (3)}$

Simplify.

$x = \frac{- 2 \pm \sqrt{- 8}}{6}$

The discriminant is negative, so this equation has no real solutions.

Example 3:

Solve the quadratic equation.

$x^{2} - 4 x + 2 = 0$

Here $a = 1, b = - 4, and c = 2$ . Substituting, we get:

$\begin{array}{l} x = \frac{- (- 4) \pm \sqrt{{(- 4)}^{2} - 4 (1) (2)}}{2 (1)} \\ = \frac{4 \pm \sqrt{16 - 8}}{2} \\ = \frac{4 \pm \sqrt{8}}{2} \end{array}$

Simplify.

$\begin{array}{l} x = \frac{4 \pm \sqrt{4 \cdot 2}}{2} \\ = \frac{4 \pm 2 \sqrt{2}}{2} \\ = \frac{2 (2 \pm \sqrt{2})}{2} \\ = 2 \pm \sqrt{2} \end{array}$

The discriminant is positive but not a perfect square, so we have two real solutions:

$x = 2 + \sqrt{2}$ and $x = 2 - \sqrt{2}$

Quadratic Formula

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