Lesson Business-related problems to solve them using quadratic equations

Business-related problems to solve them using quadratic equations

Problem 1

Let x be the number of shares he sold.

Then the number of shares he purchased was (x+6).


The selling price was   dollars per share.

The buying price was  .


According to the problem, 

     -  = 2  dollars per share.


At this point, the setup is done.

To solve the equation, multiply both sides by x*(x+6).  You will get then

    2880(x+6) - 2880x = 2x*(x+6)

    1440(x+6) - 1440x = x(x+6)

          1440*6      = x^2 + 6x

          x^2 + 6x - 8640 = 0

          (x-90)*(x+96) = 0.


The only positive root, which makes sense is x= 90.


So, the number of shares he sold was 90.     ANSWER


CHECK.  The selling price was   = 32 dollars per share.

        The buying price was   = 30 dollars per share.

        The difference in prices was  32 - 30 = 2 dollars per share.   ! Correct !

Problem 2

Let n be the number of shares Paul sold.

Then the original number of shares he bought was (n+50).


The buying price per share was   dollars.

The selling price per share was  dollars.


The difference of these prices was 5 dollars, according to the problem,
which gives this "price" equation


     -  = 5  dollars.     (1)


To solve this equation, multiply both sides by n*(n+50), then simplify step by step


    3000*(n+50) - 3000*n = 5n*(n+50)

    3000n + 150000 - 3000n = 5n^2 + 250n

    5n^2 + 250n - 150000 = 0

     n^2 +  50n -  30000 = 0

     (n+200)*(n-150) = 0


The equation has two roots, -200 and 150.  
Of these two roots, we discard the negative one and accept the positive value n= 150.


CHECK.  To check, substitute n= 150 into equation (1). You will get

         -  = 20 - 15 = 5,  which is precisely correct.


ANSWER.  Paul sold 150 shares.

Problem 3

Let x be the number of their product that were sold last year.


From [2], we have  this expression for margin profit last year 

    ,  or  .            (1)


From [3], we have  this expression for margin profit this year 

    ,  or      (2)


From [1], quantity (2) is 0.02 greater than quantity (1).  It gives this equation

     -  = 0.02.


    +-------------------------------------------------------------------+
    |    In this problem, the MAJOR STEP is to get a setup equation,    |
    |            and we successfully overcome it (!)                    |
    +-------------------------------------------------------------------+

           +---------------------------------------+
           |   To solve it is just a brute force.  |
           +---------------------------------------+



Multiply both sides by x*(x+2000)

    2040x - 1500(x + 2000) = 0.02x*(x+2000)


Simplify step by step

    2040x - 1500x - 3000000 = 0.02x^2 + 40x

    0.02x^2 - 500x + 3000000 = 0


The roots of this quadratic equation are (use the quadratic formula)

     = 10000;   = 15000.


Value  = 10000 gives the profit margin   = 0.15 = 15%,
which contradicts to condition [4]; so we reject this root.


Value  = 15000 gives the profit margin   = 0.10 = 10%,
which agrees with condition [4]; so we accept this solution.


Thus last year the company sold 15000 units of the product.

This year they sold 2000 units more, i.e.  15000 + 2000 = 17000 units of the product.