+0  
 
0
63
2
avatar+418 

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?

Logic  Sep 14, 2018
 #1
avatar
+1

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.

Guest Sep 15, 2018
 #2
avatar+90968 
+1

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

 

 

cool cool cool

CPhill  Sep 15, 2018

29 Online Users

avatar
avatar
avatar
avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.