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

1 Answer
Gió
Jul 26, 2015

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

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