SOLUTION: The area of a rectangular wall in a classroom is 308 feet. Its length is 5 feet shorter than three times its width. Find the length and width of the wall of the classroom. So

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Question 1102687: The area of a rectangular wall in a classroom is 308 feet. Its length is 5 feet shorter than three times its width. Find the length and width of the wall of the classroom.
So the equation is w(3w-5)=308, right? and that turns into 3w%5E2-5w=308. And then you complete the square? But I can seem to be able to find any real solutions. Can you help?

Thank you in advance!
Found 2 solutions by Alan3354, greenestamps:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
The area of a rectangular wall in a classroom is 308 feet. Its length is 5 feet shorter than three times its width. Find the length and width of the wall of the classroom.
So the equation is w(3w-5)=308, right? and that turns into 3w%5E2-5w=308. And then you complete the square? But I can seem to be able to find any real solutions. Can you help?
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3w^2 - 5w - 308 = 0
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 3x%5E2%2B-5x%2B-308+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%28-5%29%5E2-4%2A3%2A-308=3721.

Discriminant d=3721 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--5%2B-sqrt%28+3721+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-5%29%2Bsqrt%28+3721+%29%29%2F2%5C3+=+11
x%5B2%5D+=+%28-%28-5%29-sqrt%28+3721+%29%29%2F2%5C3+=+-9.33333333333333

Quadratic expression 3x%5E2%2B-5x%2B-308 can be factored:
3x%5E2%2B-5x%2B-308+=+%28x-11%29%2A%28x--9.33333333333333%29
Again, the answer is: 11, -9.33333333333333. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+3%2Ax%5E2%2B-5%2Ax%2B-308+%29

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11 by 28


Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


Solving a quadratic by completing the square is seldom done; it is, in general, not an efficient method.

There are problems where you need to complete a square to solve the problem; but usually a quadratic equation is solved using the quadratic formula, or a graphing calculator of some other tool, or, if the equation has integer coefficients that are relatively small, by factoring.

Your quadratic equation is
3w%5E2-5w=308,
or
3w%5E2-5w-308=0

To solve this by factoring, you need to find two integers whose product is 308, giving you the constant term in the equation.

More specifically, you need to find numbers a and b for which
%283w-a%29%28w%2Bb%29+=+3w%5E2-5w-308

After a lot of work, you might come up with the correct factorization, which is
%283w%2B28%29%28w-11%29+=+3w%5E2-5w-308

That will give you the answer: the width is w=11; the length is 3(11)-5 = 28.


But wait a minute!. This is looking like a lot of work, looking for those numbers a and b....

The original problem was to find two numbers whose product is 308 and in which the larger is 5 less than 3 times the smaller. Since solving the problem by factoring still requires us to find two numbers whose product is 308, and which satisfy conditions more complicated than the original problem, the easiest way to solve this problem is to forget the formal algebraic methods and simply look for two numbers that satisfy the conditions of the original problem.

We are looking for a pair of numbers whose product is 308, in which the larger is 5 less than 3 times the smaller. So let's write down some pairs of whole numbers whose product is 308:

154*2
77*4
44*7
28*11
14*22

28 and 11 work! They are the dimensions of the wall.