How do you solve #-5x^2-2x-4 = 0# ?

1 Answer
Nov 11, 2017

Answer:

#x = 1/5(-1+-sqrt(19)i)#

Explanation:

Complete the square, then use the difference of squares identity:

#a^2-b^2 = (a-b)(a+b)#

with #a=(5x+1)# and #b=sqrt(19)i# as follows:

#0 = -5(-5x^2-2x-4)#

#color(white)(0) = 25x^2+10x+20#

#color(white)(0) = (5x)^2+2(5x)+1+19#

#color(white)(0) = (5x+1)^2+(sqrt(19))^2#

#color(white)(0) = (5x+1)^2-(sqrt(19)i)^2#

#color(white)(0) = ((5x+1)-sqrt(19)i)((5x+1)+sqrt(19)i)#

#color(white)(0) = (5x+1-sqrt(19)i)(5x+1+sqrt(19)i)#

Hence:

#5x = -1+-sqrt(19)i#

So:

#x = 1/5(-1+-sqrt(19)i)#