SOLUTION: I am having trouble with a quadratic word problem. The word problem is as follows: The sides of one square have length 3 m more than the sides of a second square. If the area of

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Question 414895: I am having trouble with a quadratic word problem. The word problem is as follows:
The sides of one square have length 3 m more than the sides of a second square. If the area of the larger square is subtracted from 4 times the area of the smaller square, the result is 36 m^2. What are the lengths of the sides of each square?

Found 2 solutions by stanbon, josmiceli:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
The sides of one square have length 3 m more than the sides of a second square. If the area of the larger square is subtracted from 4 times the area of the smaller square, the result is 36 m^2. What are the lengths of the sides of each square?
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1st Square DATA:
Side = x
Area = x^2
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2nd Square DATA:
side = x+3
Area = (x+3)^2
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Equation:
4*smaller area - larger area = 36 m^2
4x^2 - (x+3)^2 = 36
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4x^2-[x^2+6x+9] - 36 = 0
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4x^-x^2-6x-9-36 = 0
3x^2-6x-45 = 0
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x^2-2x-15 = 0
(x-7)(x+2) = 0
Positive solution:
x = 7m (side of smaller square)
x+3 = 10m (side of larger square)
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Cheers,
Stan H.
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Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let the side of the larger square = s+%2B+3 m
The side of the smaller square = s m
given:
+4s%5E2+-+%28s+%2B+3%29%2A%28s+%2B+3%29+=+36 m2
4s%5E2+-+%28s%5E2+%2B+6s+%2B+9%29+=+36
+3s%5E2+-+6s++-+45+=+0
s%5E2+-+2s+-+15+=+0
s+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
a+=+1
b+=+-2
c+=+-15
s+=+%28-%28-2%29+%2B-+sqrt%28+%28-2%29%5E2-4%2A1%2A%28-15%29+%29%29%2F%282%2A1%29+
s+=+%28+2+%2B-+sqrt%28+4+%2B+60+%29%29+%2F+2+
s+=+%282+%2B+8%29%2F2 (can't use the negative root)
s+=+5
and
s+%2B+3+=+8
The sides are 5 m and 8 m
check answer:
+4s%5E2+-+%28s+%2B+3%29%2A%28s+%2B+3%29+=+36
+4%2A5%5E2+-+%285+%2B+3%29%2A%285+%2B+3%29+=+36
100+-+64+=+36
36+=+36
OK