How do you solve # x^2+5x+8=0#?

1 Answer
Apr 30, 2016

Answer:

#x = -5/2+-sqrt(7)/2i#

Explanation:

Pre-multiply by #4# (to avoid arithmetic with fractions), complete the square then use the difference of squares identity:

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

with #a=(2x+5)# and #b=sqrt(7)i# as follows:

#0 = 4(x^2+5x+8)#

#=4x^2+20x+32#

#=(2x)^2+2(2x)(5)+32#

#=(2x+5)^2-25+32#

#=(2x+5)^2+7#

#=(2x+5)^2-(sqrt(7)i)^2#

#=((2x+5)-sqrt(7)i)((2x+5)+sqrt(7)i)#

#=(2x+5-sqrt(7)i)(2x+5+sqrt(7)i)#

So: #2x = -5+-sqrt(7)i#

So: #x = -5/2+-sqrt(7)/2i#

#color(white)()#
Alternatively, use the quadratic formula:

The equation #x^2+5x+8 = 0# is of the form #ax^2+bx+c = 0# with #a=1#, #b=5# and #c=8#.

This has roots given by the quadratic formula:

#x = (-b+-sqrt(b^2-4ac))/(2a)#

#(-5+-sqrt(5^2-(4*1*8)))/(2*1)#

#=(-5+-sqrt(25-32))/2#

#=(-5+-sqrt(-7))/2#

#=(-5+-sqrt(7)i)/2#

#=-5/2+-sqrt(7)/2i#