How do you find the zeros, real and imaginary, of #y= -x^2-12x-8 # using the quadratic formula?

1 Answer
Jan 31, 2017

Answer:

#x=-6+2sqrt7;# or #x=-6-2sqrt7#

Explanation:

When using the quadratic formula, the zeros are the two values for #x#. When solving quadratic equations, the equation must be equal to #0#.

Your equation: #y=-x^2-12x-8#

Set #y# equal to zero.

#0=-x^2-12x-8#

Multiply by #-1# to make #a# positive.

#x^2+12x+-8#

#-x^2-12x-8# is a quadratic equation in the form of #"color(red)(a)^2+color(blue)bx+color(green)(c)#, where the coefficients are #"color(red)(a)=(1)#, #color(blue)(b)=12#, #color(green)(c)=8#.

Quadratic equation

#x=(color(blue)(-b)+-sqrt(color(blue)(b)^2-4color(red)(a)color(green)(c)))/(2color(red)a)#

Substitute the known values into the equation.

#x=("-(color(blue)(12))+-sqrt(color(blue)(12)^2-4(color(red)(1))(color(green)(8))))/((2*color(red)(1)))#

#x=(-12+-sqrt(144-(32)))/(-2)#

#=(-12+-sqrt(112))/(-2)#

Factor #sqrt112#.

#sqrt112=sqrt(16xx7)=4sqrt7#

#x=(-12+-4sqrt7)/2#

Simplifyby factoring out #2# in #-12, +-4, and 2#.

#x=-6+-2sqrt7#

#x=-6+2sqrt7;# or #x=-6-2sqrt7#