How do you find the zeros, real and imaginary, of #y=x^2+32x-4# using the quadratic formula?

1 Answer
Apr 29, 2018

Answer:

#x = -16 +- 2sqrt65#

Explanation:

#y = x^2 + 32x - 4#

To use the quadratic formula to find the zeroes, we need to make sure the equation is written in the form #color(red)(a)x^2 + color(magenta)(b)x + color(blue)(c) = 0#, which this equation is.

So we know that:
#color(red)(a = 1)#

#color(magenta)(b = 32)#

#color(blue)(c = -4)#

The quadratic formula is #x = (-color(magenta)(b) +- sqrt(color(magenta)(b)^2 - 4color(red)(a)color(blue)(c)))/(2color(red)(a))#.

Now we can plug in the values for #color(red)(a)#, #color(magenta)(b)#, and #color(blue)(c)# into the quadratic formula:

#x = (-color(magenta)(32) +- sqrt((color(magenta)(32))^2 - 4(color(red)(1))(color(blue)(-4))))/(2(color(red)(1)))#

Simplify:
#x = (-32 +- sqrt(1024 + 16))/2#

#x = (-32 +- sqrt(1040))/2#

#x = (-32 +- 4sqrt65)/2#

#x = -16 +- 2sqrt65#

This is the same thing as:
#x = -16 + 2sqrt65# and #x = -16 - 2sqrt65#
because #+-# means "plus or minus."

Hope this helps!