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

1 Answer
Feb 3, 2016

Answer:

#x=1/2+-sqrt(3)/4#

Explanation:

The quadratic formula for the zeros of the general equation
#color(white)("XXX")y=ax^2+bx+c#
is given by
#color(white)("XXX")x=(-b+-sqrt(b^2-4ac))/(2a)#

For the given equation
#color(white)("XXX")y=3x^2-32x+2#
this becomes
#color(white)("XXX")(32+-sqrt(32^2-4(32)(2)))/(2(32)#

and from there on, it's just basic arithmetic to get
#color(white)("XXX")x=1/2+-sqrt(3)/4#