Question 1174517
You and your friend part at an intersection.
 Your friend drives away north at a constant speed.
 You linger at the intersection for an hour, and then drive off due east at a speed that is 11 miles per hour faster than your friend's speed.
 2 hours after your friend's departure the distance between the two of you is 106 miles.
 Your friend is traveling at a speed of miles per hours.
:
let s = friends speed mph
since he travels 2 hrs
2s = distance friend is from intersection
:
Your travel time is only 1 hr since you started 1 hr later,
traveling 11 mph faster, therefore
1(s+11) = your travel distance
:
pythag: a^2 + b^2 = c^2, where
a = 2s
b = (s+11)
c = 106
:
(2s)^2 + (s+11)^2 = 106^2
FOIL(s+11)(s+11)
4s^2 + (s^2 + 22s + 121) = 11236
4s^2 + s^2 + 22s + 121 - 11236 = 0
Combine like terms, a quadratic equation
5s^2 + 22s - 11115 = 0
Use the quadratic formula: a=5; b=22, c=-11115
I got a positive solution of
s = 45 mph is your friends speed
:
:
Check on your calc
{{{sqrt(90^2 + 56^2)}}} = 106