How do you find the roots, real and imaginary, of #y=-(x -4 )^2-2x-1# using the quadratic formula?

1 Answer
Dec 27, 2017

Answer:

#x_1=3+2sqrt2i#
#x_2=3-2sqrt2i#

Explanation:

#y=-(x-4)^2-2x-1=#
#=-(x^2-8x+16)-2x-1=#
#=-x^2+8x-16-2x-1=#
#=-x^2+6x-17#

Let #y=0#
#=>#
#-x^2+6x-17=0 iff a=-1, b=6, c=-17#
#=>#
#x_{1,2}={-b+-sqrt{b^2-4ac}}/{2a}=#
#={-6+-sqrt{6^2-4*(-1)*(-17)}}/{2(-1)}=#
#={-6+-sqrt{36-68}}/{-2}=#
#={-6+-sqrt{-32}}/{-2}=#
#={-6+-sqrt{-(16*2)}}/{-2}=#
#={-6+-4sqrt2i}/{-2}=#
#={cancel(-2)(3+-2sqrt2i)}/{cancel(-2)}=#
#=3+-2sqrt2i#
#=>#
#x_1=3+2sqrt2i#
#x_2=3-2sqrt2i#