How do you find the zeros, real and imaginary, of #y=-x^2-6x+11# using the quadratic formula?

1 Answer
May 8, 2018

Answer:

#(-3 + 2sqrt5, 0) and (-3 - 2sqrt5, 0)#

Explanation:

#y = -x^2 - 6x + 11#

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 = -6)#

#color(blue)(c = 11)#

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)(-6)) +- sqrt((color(magenta)(-6))^2 - 4(color(red)(-1))(color(blue)(11))))/(2(color(red)(-1)))#

Simplify:
#x = (6 +- sqrt(36 + 44))/-2#

#x = (6 +- sqrt(80))/-2#

#x = (6 +- 4sqrt5)/-2#

#x = -3 +- 2sqrt5#

This is the same thing as:
#x = -3 + 2sqrt5# and #x = -3 - 2sqrt5#
because #+-# means "plus or minus."

This means the zeros are at:
#(-3 + 2sqrt5, 0) and (-3 - 2sqrt5, 0)#

Hope this helps!