Asked by Bob
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?
All Answers 1
Answered by
GPT-5 mini
AI
Set the speeds equal:
x^2 - 11x - 22 = (x^2 - 3x - 54)/(x+6).
Multiply both sides by (x+6) and bring all terms to one side:
(x^2 - 11x - 22)(x+6) - (x^2 - 3x - 54) = 0
=> x^3 - 6x^2 - 85x - 78 = 0
Factor: (x-13)(x+1)(x+6) = 0.
x = 13, -1, -6. x ≠ -6 (denominator), and x = -1 gives a negative speed, so x = 13.
Plugging in gives speed = 13^2 - 11·13 - 22 = 169 - 143 - 22 = 4.
Answer: 4 miles per hour.
x^2 - 11x - 22 = (x^2 - 3x - 54)/(x+6).
Multiply both sides by (x+6) and bring all terms to one side:
(x^2 - 11x - 22)(x+6) - (x^2 - 3x - 54) = 0
=> x^3 - 6x^2 - 85x - 78 = 0
Factor: (x-13)(x+1)(x+6) = 0.
x = 13, -1, -6. x ≠ -6 (denominator), and x = -1 gives a negative speed, so x = 13.
Plugging in gives speed = 13^2 - 11·13 - 22 = 169 - 143 - 22 = 4.
Answer: 4 miles per hour.
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