SOLUTION: The length of a rectangle is 4 cm more than 2 times its width. If the area of the rectangle is 74 cm2, find the dimensions of the rectangle to the nearest thousandth. The answe

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: The length of a rectangle is 4 cm more than 2 times its width. If the area of the rectangle is 74 cm2, find the dimensions of the rectangle to the nearest thousandth. The answe      Log On


   

Question 69253: The length of a rectangle is 4 cm more than 2 times its width. If the area of the rectangle is 74 cm2, find the dimensions of the rectangle to the nearest thousandth.
The answer that I have figured is w=5.165 and L=14.330 and I have no idea if this is correct. Can I get a little help with this problem. Thanks.
Answer by ptaylor(2198) About Me  (Show Source):
You can put this solution on YOUR website!
I WOULD SAY THAT YOU ARE RIGHT ON THE MONEY ON THIS PROBLEM. GOOD WORK!!!!

Let x= the width of the rectangle
Then 2x+4=length of rectangle (4 cm more than twice the width)
Area of rectangle =l*w=(2x+4)(x) so our equation to solve is:
%282x%2B4%29%28x%29=74 get rid of parens
2x%5E2%2B4x=74 subtract 74 from both sides

2x%5E2%2B4x-74=0 divide both sides by 2
x%5E2%2B2x-37=0 quadratic in standard form
We'll solve using the quadratic formula:
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
x+=+%28-2+%2B-+sqrt%28+2%5E2-4%2A%28-37%29+%29%29%2F%282%29+
x+=+%28-2+%2B-+sqrt%28152+%29%29%2F%282%29+
x+=+%28-2+%2B-+sqrt%28152+%29%29%2F%282%29+
x+=+%28-2+%2B-12.330+%29%29%2F%282%29+
x+=+%28-2+%2B12.330+%29%29%2F%282%29+
x+=+%2810.330+%29%29%2F%282%29+
x=5.165 cm -----------------width of rectangle
2x%2B4=2%2A5.165%2B4=14.330 cm---------------length of rectangle
We'll discount the negative value for x
CK
Area = l*w=14.330*5.165=74.01~74 ---ok
Hope this helps----ptaylor