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

2 Answers
Jul 12, 2018

Answer:

The roots are at x = -28, 29.

Explanation:

#y = x(x-1)-812#

Before we use the quadratic formula, we have to make the equation in standard quadratic form, or #y = ax^2 + bx + c#.

Let's distribute the #x(x-1)#:
#y = x^2 - x - 812#

As you can see, this is now in the form #y = ax^2 + bx + c#, where #color(red)(a = 1)#, #color(magenta)(b = -1)#, and #color(blue)(c = -812)#.

The quadratic equation is used to find the real and imaginary roots/zeros of a quadratic equation. It's formula is #x = (-color(magenta)(b) +- sqrt((color(magenta)(b))^2-4color(red)(a)color(blue)(c)))/(2color(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 = (1 +- sqrt(1 + 3248))/2#

#x = (1+- sqrt(3249))/2#

#x = (1+-57)/2#

#x = (1+57)/2#, #x = (1-57)/2#

#x = 58/2#, #x = -56/2#

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

To show that this is correct, I graphed the original equation #y = x(x-1) - 812# using desmos.com:
enter image source here

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

Answer:

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

Explanation:

#y = x^2 - x - 812#

Discriminant # D = (b^2 - 4ac)#

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 # = (-b +- sqrt (d) ) / (2a)#

#x = (1 +- sqrt 3249) / 2 = (1 +- 57) / 2 = 29, -28#