How do you find the zeros of # y = -80x^2 +32x + 16 # using the quadratic formula?

1 Answer
Nov 30, 2015

Answer:

You just need to "plug and chug" - put the values into the equation and solve for x.

Explanation:

The Quadratic Formula uses the "a", "b", and "c" from "ax2 + bx + c", where "a", "b", and "c" are just numbers. The Formula is derived from the process of completing the square, and is formally stated as:
For #ax^2 + bx + c = 0#, the value of x is given by:

#x = (-b +- sqrt[b^2 – 4ac])/(2a)#

Note that, for the Formula to work, you must have "(quadratic) = 0". Note also that the "2a" at the bottom of the Formula is underneath everything above, not just the square root. And don't forget that it's a "2a" under there, not just a "2"! And make sure that you are careful not to drop the square root or the "plus/minus" in the middle of your calculations, or I can guarantee that you will forget to "put them back" on your test, and you'll mess yourself up.

And remember that "b2" means "the square of ALL of b, including the sign", so don't leave b2 being negative, even if b is negative, because the square of a negative is a positive. In other words, don't be sloppy and don't try to take shortcuts, because it will only hurt you in the long run. Trust me on this!

From: Stapel, Elizabeth. "The Quadratic Formula: The Discriminant and Graphs." Purplemath.
Available from http://www.purplemath.com/modules/quadform3.htm.
Accessed 05 July 2007