# What is the quadratic formula used for?

Oct 19, 2014

The quadratic formula is used to get the roots of a quadratic equation, if the roots exists at all.

We usually just perform factorization to get the roots of a quadratic equation. However, this is not always possible (especially when the roots are irrational)

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

Example 1:

$y = {x}^{2} - 3 x - 4$
$0 = {x}^{2} - 3 x - 4$

$\implies 0 = \left(x - 4\right) \left(x + 1\right)$
$\implies x = 4 , x = - 1$

Using the quadratic formula, let's try to solve the same equation

$x = \frac{- \left(- 3\right) \pm \sqrt{{\left(- 3\right)}^{2} - 4 \cdot 1 \cdot \left(- 4\right)}}{2 \cdot 1}$
$\implies x = \frac{3 \pm \sqrt{9 + 16}}{2}$
$\implies x = \frac{3 \pm \sqrt{25}}{2}$
$\implies x = \frac{3 + 5}{2} , x = \frac{3 - 5}{2}$
$\implies x = 4 , x = - 1$

Example 2:

$y = 2 {x}^{2} - 3 x - 5$
$0 = 2 {x}^{2} - 3 x - 5$

Performing factorization is a little hard for this equation, so let's jump straight to using the quadratic formula

$x = \frac{- \left(- 3\right) \pm \sqrt{{\left(- 3\right)}^{2} - 4 \cdot 2 \cdot \left(- 5\right)}}{2 \cdot 2}$

$x = \frac{3 \pm \sqrt{9 + 40}}{4}$

$x = \frac{3 \pm \sqrt{49}}{4}$

$x = \frac{3 + 7}{4} , x = \frac{3 - 7}{4}$

$x = \frac{5}{2} , x = - 1$