# How do you find the roots, real and imaginary, of y=(x-1)(x+1) using the quadratic formula?

Dec 19, 2017

$x = \pm 1$

#### Explanation:

$y = \left(x - 1\right) \left(x + 1\right)$

The roots of the equation occur when $y = 0$

i.e. where $\left(x - 1\right) \left(x + 1\right) = 0$

This is clearly $x = \pm 1$

However, in this question we are asked to use the quadratic formula which states that where $a {x}^{2} + b x + c = 0$ then:

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

In our case; $\left(x - 1\right) \left(x + 1\right) = 0$

$\to {x}^{2} - 1 = 0$

Hence, $a = 1 , b = 0 , c = - 1$

$x = \frac{- 0 \pm \sqrt{{0}^{2} - 4 \cdot 1 \cdot \left(- 1\right)}}{2 \cdot 1}$

$= \frac{\pm \sqrt{4}}{2} = \pm \frac{2}{2}$

$= \pm 1$