SOLUTION: The length of a rectangle is 13 centimeters less than four times its width. Its area is 35 square centimeters. Find the dimensions of the rectangle.

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Question 1056260: The length of a rectangle is 13 centimeters less than four times its width. Its area is 35 square centimeters. Find the dimensions of the rectangle.
Found 2 solutions by addingup, MathTherapy:
Answer by addingup(3677)   (Show Source): You can put this solution on YOUR website!
L*W = 35
L = 4W-13
rewrite, substituting for L:
(4W-13)*W = 35
I don't know if you have to solve by completing the squares or by using the quadratic formula. I'm going to complete the squares. Send me an email if you need to use the quadratic formula and I'll send it to you.
:
Expand, distribute the W onto the terms inside the ():
W*4W+(-13)W = 35
W^2*4+(-13)W = 35
W^2*4+(-13)W-35 = 0 Now add 35 to both sides, like this:
4W*2+(-13)W-35-(-35 = -(-35) and we get:
4W^+(-13)W = 35 get rid of the 4 in front, divide both sides by 4
W^2+(-13/4)W = 35/4
Now to complete the squares: The coefficient of W is -13/4, as you see in the equation. Divide it by 2: (-13/4)/2/1 = -13/4*1/2 = (-13*1)/(4*2) = -13/8 Now ad the square of this number to both sides of the equation, add (-13/8)^2= 169/64 to both sides:
W^2+(-13/4)W+169/64 = 729/64 factor the left:
(W-13/8)^2 = 729/64
sqrt[(W-13/8)^2] = sqrt(729/64
= W-13/8 = 27/8 or W-13/8 = -27/8
W = 5 or W = -7/4
Let's try the 5 first:
W = 5
L / (4*5)-13 = 7
Area: L*W = 7*5 = 35 Correct




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

The length of a rectangle is 13 centimeters less than four times its width. Its area is 35 square centimeters. Find the dimensions of the rectangle.
Let width be W

Then length = 4W - 13

We then get the following AREA equation: W(4W - 13) = 35





(W - 5)(4W + 7) = 0----- Factoring the TRINOMIAL

W, or width =          OR         (ignore)

Length:  

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