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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?

 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.

 Sep 15, 2018
 #2
avatar+98042 
+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

 Sep 15, 2018

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