Given r^2-9r-38=-9r2−9r−38=−9
First add 9 to both sides to get:
r^2-9r-29=0r2−9r−29=0
Then
0 = r^2-9r-290=r2−9r−29
= r^2-9r+81/4-81/4-29=r2−9r+814−814−29
= (r-9/2)^2-(81/4+29)=(r−92)2−(814+29)
= (r-9/2)^2-(81/4+116/4)=(r−92)2−(814+1164)
= (r-9/2)^2-197/4=(r−92)2−1974
Adding 197/41974 to both ends of this equation, we find:
(r-9/2)^2 = 197/4(r−92)2=1974
So:
r-9/2 = +-sqrt(197/4) = +-sqrt(197)/sqrt(4) = +-sqrt(197)/2r−92=±√1974=±√197√4=±√1972
Add 9/292 to both sides to get:
r = 9/2+-sqrt(197)/2 = (9+-sqrt(197))/2r=92±√1972=9±√1972
In general:
ar^2+br+c = a(r+b/(2a))^2 + (c - b^2/(4a))ar2+br+c=a(r+b2a)2+(c−b24a)
which is zero when
a(r+b/(2a))^2 = (b^2/(4a) - c) = (b^2 - 4ac)/(4a)a(r+b2a)2=(b24a−c)=b2−4ac4a
and hence
(r+b/(2a))^2 = (b^2 - 4ac)/(4a^2)(r+b2a)2=b2−4ac4a2
so
r+b/(2a) = +-sqrt((b^2 - 4ac)/(4a^2)) = +-sqrt(b^2 - 4ac)/sqrt(4a^2)r+b2a=±√b2−4ac4a2=±√b2−4ac√4a2
= +-sqrt(b^2-4ac)/(2a)=±√b2−4ac2a
Subtracting b/(2a)b2a from both sides we get:
r = -b/(2a)+-sqrt(b^2-4ac)/(2a) = (-b +- sqrt(b^2-4ac))/(2a)r=−b2a±√b2−4ac2a=−b±√b2−4ac2a
Does that look familiar?
