SOLUTION: The length of a rectangle is 3 less than twice the width. If the area of the rectangle is 25in^2, find the length of each side to the nearest hundredth.

Algebra ->  Rectangles -> SOLUTION: The length of a rectangle is 3 less than twice the width. If the area of the rectangle is 25in^2, find the length of each side to the nearest hundredth.      Log On


   

Question 775473: The length of a rectangle is 3 less than twice the width. If the area of the rectangle is 25in^2, find the length of each side to the nearest hundredth.
Answer by sofiyac(983) About Me  (Show Source):
You can put this solution on YOUR website!
let the width be x, then the length is 2x-3 area is width times length so
x(2x-3)=25 solve for x
2x^2-3x-25=0
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
x+=+%283+%2B-+sqrt%28+%28-3%29%5E2-4%2A2%2A%28-25%29+%29%29%2F%282%2A2%29+
x+=+%283+%2B-+sqrt%28+9%2B200+%29%29%2F%284%29+
x+=+%283+%2B-+sqrt%28+209+%29%29%2F%284%29+
x+=+%283+%2B-+14.46%29%2F%284%29+
x+=+%283+%2B+14.46%29%2F%284%29+
x=4.37
x+=+%283+-+14.46%29%2F%284%29+
x=-2.87 obviously we can't have a negative width so we're just going with the first answer so, if width is 4.37 then length would be 2(4.37)-3=5.74