SOLUTION: A baseball court is 2 m longer than it is wide. If its area is 575 m^2 find the dimensions of the court.

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Question 1160886: A baseball court is 2 m longer than it is wide. If its area is 575 m^2 find the dimensions of the court.
Found 4 solutions by MathLover1, MathTherapy, greenestamps, saw:
Answer by MathLover1(20850) About Me  (Show Source):
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if a baseball court is 2m longer than it is wide we have
L=W%2B2m .........eq.1
if its area is 575m%5E2, we have
L%2AW=575m%5E2........eq.2
substitute L from eq.1 in eq.2
%28W%2B2m%29%2AW=575m%5E2............solve for W
W%5E2%2B2m%2AW=575m%5E2
W%5E2%2B2m%2AW-575m%5E2=0...factor completely
W%5E2-23m%2AW%2B25mW-575m%5E2=0
%28W%5E2-23m%2AW%29%2B%2825mW-575m%5E2%29=0
W%28W-23m%29%2B25m%28W-23m%29=0
%28W+-+23m%29+%28W+%2B+25m%29+=+0

=> %28W+-+23m%29+=+0->W+=+23m
=> %28W+%2B25m%29+=+0->W+=+-25m->disregard negative solution


go to L=W%2B2m .........eq.1, substitute W

L=23m%2B2m
L=25m

so, the dimensions of the court are:
the length: 25m
the width: 23m



Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!

A baseball court is 2 m longer than it is wide. If its area is 575 m^2 find the dimensions of the court.
No need to "write a book!"

All that's needed here are 2 integers that have a product of 575.

These are 23 and 25.

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


You should know HOW to solve a problem like this using formal algebra....

HOWEVER, note that in this problem (as shown by one of the solutions you have received), like in many simple problems, using formal algebra has you end up factoring a quadratic equation by finding two numbers whose difference is 2 and whose product is 575. But that's exactly what the original problem asks you to do -- so formal algebra only adds extra work to finding the answer; it doesn't make solving the problem any easier.

So do as tutor #MathTherapy says: simply find two numbers whose product is 575 and whose difference is 2.

You should recognize that 575 is a multiple of 25, and 575/25 = 23. And 23 and 25 satisfy the required conditions, so you are done.

Answer by saw(34) About Me  (Show Source):
You can put this solution on YOUR website!
Solutions.
Let length = l
Let width = w
As the condition given
l = w+2m
l*w = 575m^2
Use substitution
(w+2m)*w = 575m^2
w^2+2mw = 575m^2
w^2+2mw-575m^2 = 0
(w-23m)(w+25m) = 0
w-23m = 0 or w+25m = 0
w = 23m or w = -25m (negative is not possible)
w = 23m
l = w+2m
Use substitution
l = 23m+2m
l = 25m
Check-----
l*w = 575m^2