SOLUTION: The area of a rectangle is 32cm^2. The width of the rectangle is 4 cm more than 3 times the length. What is the length of the rectangle is cm? Thanks in advance for any hel

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons -> SOLUTION: The area of a rectangle is 32cm^2. The width of the rectangle is 4 cm more than 3 times the length. What is the length of the rectangle is cm? Thanks in advance for any hel      Log On


   

Question 1041079: The area of a rectangle is 32cm^2. The width of the rectangle is 4 cm more than 3 times the length.
What is the length of the rectangle is cm?

Thanks in advance for any help, would love any tips on translating the word problem into an equation.
Found 2 solutions by Boreal, stanbon:
Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
length is smaller and call it L
width is 4 more than 3 times length. That is 4+3L
The area, product, is 32 cm^2
L(4+3L)=32
4L+3L^2=32
3L^2+4L-32=0, just by rewriting and setting equal to 0.
(3L-8)(L+4)=0. This is guess and check factoring. I need a product of -32 and 8 and 4 are factors that might work. If not, check 16 and 2
L= -4, can't work
3L=8;L=8/3 cm
3(8/3)+4; and 4+8=12 cm
The width is 12 cm
The length is 8/3 cm ANSWER.
Their product is 32 cm^2.

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
The area of a rectangle is 32cm^2. The width of the rectangle is 4 cm more than 3 times the length.
What is the length of the rectangle is cm?
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Equations:
W = 3L + 4
W*L = 32 cm^2
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Substitute for "W" and solve for "L"::
L(3L+4) = 32
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3L^2 + 4L - 32 = 0
Factor::
(3L-8)(L+4) = 0
Positive solution:
L = 8/3 = 2.6666.. cm
Solve for "W:
W = 3(8/3)+4 = 12 cm
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Cheers,
Stan H.
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