Word Problem using Quadratic Equation

Question

Maria and Stanley leave Stanley's house at the same time. Maria drives north and Stanley drives west. Maria's average speed is 9 mph slower than Stanley's. At the end of one hour, they are 85 miles apart. Find Stanley's average speed. (Round your answer to the nearest tenth.)

Answer 1

What a fun problem. In tackling this problem, I visualized what was occurring as Maria and Stanley were driving away from each other. If one thinks about it, what does the graphic representation of their routes look like? Maria drives north, while Stanley heads west. That's a 90 degree angle these two form, leaving the house at the same time.  Since we can assume their speed is uniform, and their rates do not change, we have in essence the formation of a right triangle where the hypotenuse is given to us. The hypotenuse, in this case is 85 miles. Since the two of them are traveling at an average speed, the two of them will traverse a distance in proportion to each other within an hour. Let's represent Stanley's distance as x, while Maria's distance will always be 9 mph slower, or 9 miles slower in one hours time.  Hence, Maria's distance can be represented as x -9. Now we have all three sides of a right triangle.

 

Our next objective would be to solve for x. What formula allows us to do this? The Pythagorean theorem, 

 

      a^2 + b^2 = c^2, where a, b, and c (the hypothenuse) are the three sides.

 

For our exercise, we have 

 

     x^2 + (x - 9)^2 = 85^2.

 

Expanding, we get,

 

    x^2 + (x - 9)(x - 9) = 7225.

 

Solving further, we have

 

    x^2 + x^2 - 18x + 81 = 7225,

 

and then,

 

    x^2 + x^2 - 18x -7144 = 0 (adding -7225 to both sides of the equation).

 

Now we have,

 

    2x^2 - 18x - 7144 = 0 (adding like powers).

 

Using the quadratic formula, we will get a positive and a negative value. We can disregard the negative value since we are dealing with a 'positive' distance. 

 

    Our positive value (Stanley's average speed) will be 64.4 mph!