# How do you find the roots, real and imaginary, of y=x(x-1)-812  using the quadratic formula?

Jul 12, 2018

The roots are at x = -28, 29.

#### Explanation:

$y = x \left(x - 1\right) - 812$

Before we use the quadratic formula, we have to make the equation in standard quadratic form, or $y = a {x}^{2} + b x + c$.

Let's distribute the $x \left(x - 1\right)$:
$y = {x}^{2} - x - 812$

As you can see, this is now in the form $y = a {x}^{2} + b x + c$, where $\textcolor{red}{a = 1}$, $\textcolor{m a \ge n t a}{b = - 1}$, and $\textcolor{b l u e}{c = - 812}$.

The quadratic equation is used to find the real and imaginary roots/zeros of a quadratic equation. It's formula is $x = \frac{- \textcolor{m a \ge n t a}{b} \pm \sqrt{{\left(\textcolor{m a \ge n t a}{b}\right)}^{2} - 4 \textcolor{red}{a} \textcolor{b l u e}{c}}}{2 \textcolor{red}{a}}$

Let's plug in our values into the formula and solve for $x$:
x = (-(color(magenta)(-1)) +- sqrt((color(magenta)(-1))^2 - 4(color(red)(1))(color(blue)(-812))))/(2(color(red)(1))

Now simplify:
$x = \frac{1 \pm \sqrt{1 + 3248}}{2}$

$x = \frac{1 \pm \sqrt{3249}}{2}$

$x = \frac{1 \pm 57}{2}$

$x = \frac{1 + 57}{2}$, $x = \frac{1 - 57}{2}$

$x = \frac{58}{2}$, $x = - \frac{56}{2}$

Therefore, the roots are at:
x = -28, 29

To show that this is correct, I graphed the original equation $y = x \left(x - 1\right) - 812$ using desmos.com: As you can see, the roots are indeed at $x = - 28$ and $x = 29$.

If you need another example, feel free to watch this Khan Academy video:

Hope this helps!

Jul 12, 2018

Roots are color(violet)(x = 29, -28

#### Explanation:

$y = {x}^{2} - x - 812$

Discriminant $D = \left({b}^{2} - 4 a c\right)$

If D is positive, only real root.

If it’s 0, it’s a perfect square.

If it’s negative, roots imaginary.

In our case, a = 1, b = -1, c = -812.

Hence, D = -1^2 - (4 * 1 * -812) = 3249#

Therefore, both roots are real.

Quadratic roots formula $= \frac{- b \pm \sqrt{d}}{2 a}$

$x = \frac{1 \pm \sqrt{3249}}{2} = \frac{1 \pm 57}{2} = 29 , - 28$