SOLUTION: This problem is on a quadratic equation assignment and I do not know how to set it up. The length of a rectangle is 3 cm more than 2 times its width. If the area of the rectang

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Question 68904: This problem is on a quadratic equation assignment and I do not know how to set it up.
The length of a rectangle is 3 cm more than 2 times its width. If the area of the rectangle is 99 cm^2, find both dimensions of the rectangle to the nearest thousandth.
Can someone please give me a little guidance on this problem????
Answer by rmromero(383) About Me  (Show Source):
You can put this solution on YOUR website!


What is asked in the problem?

Find the lenght and the width of the rectangle to the nearest thousandth.



Given:

Lenght of a rectangle is 3cm more than 2 times its width.

the area of the rectangle is 99 cm^2.



Representation:

Let     x = width of the rectangle

   2x + 3 = length of the rectangle



Write and equation: 

Remember that the Area of the rectangle = Length times the width



 A = (2x + 3)(x)

99 = 2x^2 + 3x

 0 = 2x^2 + 3x - 99



Factor : Use Quadratic Formula

    ax^2 + bx + c = 0, a = 2, b = 3, c= -99



       x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+

       x+=+%28-3+%2B-+sqrt%28+3%5E2-4%2A2%2A%2899%29%29%29%2F%282%2A2%29+ 

       x+=+%28-3+%2B-+sqrt%28+9+%2B+792+%29%29%2F%284%29 

       x+=+%28-3+%2B-+sqrt%28+801%29%29%2F%284%29 

       x+=+%28-3%2F4%29+%2B-+%28sqrt%28+801%29%2F4%29+



x = 6.325   and     x = -7.825



no lenght or width that is negative so disregard x = -7.825



width of the rectangle = 6.325 cm   



lenght = 2x + 3 

       = 2(6.325) + 3 

       = 12.65 + 3

       = 15.65 cm