# What is an example of using the quadratic formula?

Nov 2, 2014

Suppose that you have a function represented by $f \left(x\right) = A {x}^{2} + B x + C$.

We can use the quadratic formula to find the zeroes of this function, by setting $f \left(x\right) = A {x}^{2} + B x + C = 0$.

Technically we can also find complex roots for it, but typically one will be asked to work only with real roots. The quadratic formula is represented as:

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

... where x represents the x-coordinate of the zero.

If ${B}^{2} - 4 A C < 0$, we will be dealing with complex roots, and if ${B}^{2} - 4 A C \ge 0$, we will have real roots.

As an example, consider the function ${x}^{2} - 13 x + 12$. Here,

$A = 1 , B = - 13 , C = 12.$

Then for the quadratic formula we would have:

$x = \frac{13 \pm \sqrt{{\left(- 13\right)}^{2} - 4 \left(1\right) \left(12\right)}}{2 \left(1\right)}$ =

$\frac{13 \pm \sqrt{169 - 48}}{2} = \frac{13 \pm 11}{2}$

Thus, our roots are $x = 1$ and $x = 12$.

For an example with complex roots, we have the function $f \left(x\right) = {x}^{2} + 1$. Here $A = 1 , B = 0 , C = 1.$

$x = \frac{0 \pm \sqrt{{0}^{2} - 4 \left(1\right) \left(1\right)}}{2 \left(1\right)} = \pm \frac{\sqrt{- 4}}{2} = \pm i$
... where $i$ is the imaginary unit, defined by its property of ${i}^{2} = - 1$.