SOLUTION: The area of a playground is 168 𝑦𝑑2. The width of the playground is 2 𝑦𝑑 longer than its length. Find the length and width of the playground.

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Question 1114953: The area of a playground is 168 𝑦𝑑2.
The width of the playground is 2 𝑦𝑑 longer than its length. Find the length and width of the playground.

Found 3 solutions by Alan3354, Shin123, ikleyn:
Answer by Alan3354(69443)   (Show Source): You can put this solution on YOUR website!
Find a pair of factors of 168 that differ by 2.
==========
1*168 NG
2*84 NG
3*56 NG
etc

Answer by Shin123(626)   (Show Source): You can put this solution on YOUR website!
You have to find 2 numbers which has a product of 168 and has a difference of 2.



.......




Answer by ikleyn(52778)   (Show Source): You can put this solution on YOUR website!
.
Let x be the number (now unknown) which is exactly half way between the width and the length of the playground.


Then, as you understand,  the length = (x+1)  and  the width = (x-1).


The area = 168 = length*width = (x+1)*(x-1).


Thus (x+1)*(x-1) = 168,  which implies   = 168,  i.e.   = 168+1 = 169.


Hence, x =  = 13.


Then the length = 13+1 = 14  and the width = 13-1 = 12.

Solved.   //   This problem is for MENTAL solution.

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To see many other problems solved by same  (quite unexpected)  method,  look into the lesson
    - HOW TO solve the problem on quadratic equation mentally and avoid boring calculations
in this site.

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The method above you may find in some recreational books.

Sometimes in math circles the teachers explain it to students.


But in the school Math, the standard method of solution is to reduce the problem to a quadratic equation

w*(w+2) = 168

and then solve it using the quadratic formula or factoring.


When you do it by factoring, you need to guess two numbers with the difference of 2 and the product of 168.


You can do it by sorting through different factors of the number 168, as the tutor @shin123 explained it in his post.

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So,  it was as if you visited a Math circle today . . .



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