How do you solve #r^2+3r-3=0# by completing the square?

1 Answer
Jul 26, 2015

Answer:

I found:
#r_1=(-3+sqrt(21))/2#
#r_2=(-3-sqrt(21))/2#

Explanation:

Write it as:
#r^2+3r=3#
add and subtract #9/4# (i.e., a number that comes from #3/2# so that squared gives you #9/4# and multiplied by #2# gives you #3# which takes care of #3r#) to get:
#r^2+3rcolor(red)(+9/4-9/4)=3# rearranging:
#r^2+3r+9/4=3+9/4#
#(r+3/2)^2=(12+9)/4#
#(r+3/2)^2=21/4# square root both sides:
#r+3/2=+-sqrt(21/4)# you get two solutions:
#r_1=(-3+sqrt(21))/2#
#r_2=(-3-sqrt(21))/2#