+1206 Jack walked up a hill at a speed of \((x^2-11x-22)\) miles per hour. Meanwhile, Jill walked a total distance of \((x^2-3x-54)\) miles in \((x+6)\) hours. If Jack and Jill walked at the same speed, what is that speed, in miles per hour?
Distance = Speed x Time
Solve for x:
(x + 6) (x^2 - 11 x - 22) = x^2 - 3 x - 54
Expand out terms of the left hand side:
x^3 - 5 x^2 - 88 x - 132 = x^2 - 3 x - 54
Subtract x^2 - 3 x - 54 from both sides:
x^3 - 6 x^2 - 85 x - 78 = 0
The left hand side factors into a product with three terms:
(x - 13) (x + 1) (x + 6) = 0
Split into three equations:
x - 13 = 0 or x + 1 = 0 or x + 6 = 0
Add 13 to both sides:
x = 13 or x + 1 = 0 or x + 6 = 0
Subtract 1 from both sides:
x = 13 or x = -1 or x + 6 = 0
Subtract 6 from both sides:
x = 13 sub this in Jack's speed:(13^2 - (11*13) - 22) = 4 miles per hour.
Sub 4 into Jill's distance: (13^2 - (3*13) - 54) =76 miles. Divide this by her time:
76 / (13+6) =76/19 =4 miles per hour - Jill's speed = Jack's speed.
+128407 Distance / Rate = Speed...so
Jack's speed = Jill's speed
x^2 -11x - 22 = (x^2 - 3x - 54) / (x + 6)
x^2 - 11x - 22 = (x - 9)(x + 6) / (x + 6)
x^2 - 11x - 22 = x - 9 rearrange as
x^2 - 12x - 13 = 0 factor
(x - 13) ( x + 1) = 0
Set each factor to 0 and solve for x and we get that x = 13 or x = -1
Reject the second answer since itleads to a negative speed
If x = 13, the speed is 4 mph
