# How do you find the roots, real and imaginary, of y= (x + 1) (x - 4)  using the quadratic formula?

May 27, 2016

Roots are 4 and -1.

#### Explanation:

The roots are those numbers that assigned to the $x$ give as result zero. Because in this equation you have the multiplication of two quantities $\left(x + 1\right)$ and $\left(x - 4\right)$ it is enough that one is zero to have the roots.

$\left(x + 1\right)$ is equal to zero when $x = - 1$ and $\left(x - 4\right)$ is equal to zero when $x = 4$. So your roots are $- 1 , 4$.

But you ask about to apply the quadratic formula, To do that you have to multyply the two terms together:

$\left(x + 1\right) \left(x - 4\right) = {x}^{2} - 4 x + x - 4 = {x}^{2} - 3 x - 4$.

Now we have the equation in the format

$a {x}^{2} + b x + c$ where $a = 1$, $b = - 3$ and $c = - 4$.

The formula is: $x = \frac{- b \setminus \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$ where you can substitute your numbers
$x = \frac{3 \setminus \pm \sqrt{{\left(- 3\right)}^{2} - 4 \cdot 1 \cdot \left(- 4\right)}}{2 \cdot 1} = \frac{3 \setminus \pm \sqrt{9 + 16}}{2} = \frac{3 \setminus \pm 5}{2}$.
The solutin with the $+$ gives $4$, the one with the $-$ gives $- 1$ as we expected from the initial consideration.