# How do you find the roots, real and imaginary, of y= x^2-x-2  using the quadratic formula?

Sep 2, 2016

The roots are rational (i.e. real) and are $2$ and $- 1$.

#### Explanation:

The quadratic formula gives the roots of the general form of quadratic equation $a {x}^{2} + b x + c = 0$ as $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$.

The roots of a quadratic equation thus critically depend on the discriminant ${b}^{2} - 4 a c$.

In $y = {x}^{2} - x - 2$, the discriminant is

(-1)^2-4×1×(-2)

= $1 + 8 = 9$

As it is a complete square one can easily take square root and roots of equation will be rational (i.e. real).

The roots are

(-(-1)+-sqrt((-1)^2-4×1×(-2)))/2 or

$\frac{1 \pm \sqrt{9}}{2} = \frac{1 \pm 3}{2}$

i.e. $2$ and $- 1$