SOLUTION: A school yard had a length that was 20 more than its width. If it's area was 5376 m^2, what was the length and width of the field?

Question 1098166: A school yard had a length that was 20 more than its width. If it's area was 5376 m^2, what was the length and width of the field?
Found 2 solutions by greenestamps, ikleyn:
Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!

In the beginning, when we read this problem, we recognize that we need to find two numbers whose difference is 20 and whose product it 5376. Since it looks like that would be hard, let's see what algebra could do to help us.

Let x be the width, so x+20 is the length. Then

If we try to solve this by factoring, what we need to do is find two numbers whose difference is 20 and whose product is 5376.

BUT THAT'S WHAT THE PROBLEM TOLD US TO START WITH!!!

So using algebra didn't get us any closer to the solution....

Well... we can still use the quadratic formula to solve for x:

<-- some ugly arithmetic there...

And now we have to use a calculator to find that square root...
[obviously for this problem we want the positive root]

So the width x is 64 and the length x+20 is 84.
Check: 64*84 = 5376

That was a lot of work....

So let's go back and look at the original problem again and see if we can get to this answer with a lot less work by using logical analysis.

Once again, we need to find two numbers whose difference is 20 and whose product is 5376.

Let's think about what kinds of numbers we get when we multiply two numbers whose difference is 20. The difference of 20 means the units digits of the two numbers are the same. The units digit of our product is 6. So what single digits can we multiply by themselves to get a units digit of 6? There are two possibilities: 4 or 6.

Next think about the whole product, 5376. The square root of 5376 is 70-something, because 70 squared is 4900 and 80 squared is 6400.

So if we want to multiply two numbers whose difference is 20 and whose product is 5376, then the two numbers have to be 60-something and 80-something.

Now we just use trial and error:
66*86 = 5676 nope...
64*84 = 5376 YES!!

So by logical analysis and a bit of estimation, we have found the solution to the problem -- and with a lot less work than using formal algebra.
Answer by ikleyn(52786)   (Show Source): You can put this solution on YOUR website!
.

Let "x" be the AVERAGE of the length and the width, so that the Length = x+10 and the Width = x-10. Then the area is Area = L*W = (x+10)*(x-10) = x^2 - 100 = 5376. So, x^2 = 5376 + 100 = 5476. Hense, x = = 74. Then the Length = 74 + 10 = 84 and the Width = 74 - 10 = 64.

It is practically MENTAL solution.   ( But you must to know the T R I C K ! )

See my lesson
    - HOW TO solve the problem on quadratic equation mentally and avoid boring calculations
in this site.

--------------
For your info: there is a bunch of lessons on solving quadratic equations in this site:
    - Introduction into Quadratic Equations
    - PROOF of quadratic formula by completing the square
    - HOW TO complete the square - Learning by examples
    - HOW TO solve quadratic equation by completing the square - Learning by examples
    - Solving quadratic equations without quadratic formula
    - Who is who in quadratic equations
    - Using Vieta's theorem to solve qudratic equations and related problems
    - Using quadratic equations to solve word problems
    - HOW TO solve the problem on quadratic equation mentally and avoid boring calculations
    - OVERVIEW of lessons on solving quadratic equations

Also, you have this free of charge online textbook in ALGEBRA-I in this site
    - ALGEBRA-I - YOUR ONLINE TEXTBOOK.

The referred lessons are the part of this textbook under the topic "Quadratic equations".

Save the link to this online textbook together with its description

Free of charge online textbook in ALGEBRA-I
https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson

to your archive and use it when it is needed.

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