SOLUTION: The length of a rectangle exceeds twice its width by 3inches. If the area is 10 square inches, find the rectangle's dimensions. Round to the nearest tenth of an inch.

Algebra ->  Customizable Word Problem Solvers  -> Geometry -> SOLUTION: The length of a rectangle exceeds twice its width by 3inches. If the area is 10 square inches, find the rectangle's dimensions. Round to the nearest tenth of an inch.      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 79666: The length of a rectangle exceeds twice its width by 3inches. If the area is 10 square inches, find the rectangle's dimensions. Round to the nearest tenth of an inch.
Answer by ptaylor(2198) About Me  (Show Source):
You can put this solution on YOUR website!
Let x=width
Then 2x+3=length
Area of a rectangle =Length times Width or A=L*W
So our equation to solve is:
10=x(2x+3) get rid of parens
10=2x^2+3x subtract 10 from both sides and rearrange
2x^2+3x-10=0 quadratic in standard form; solve using the quadratic formula
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
x+=+%28-3+%2B-+sqrt%28+9%2B80+%29%29%2F%284%29+
x+=+%28-3+%2B-+sqrt%28+89+%29%29%2F%284%29+
x+=+%28-3+%2B-+9.433%29%2F%284%29+
x+=+%28-12.433%29%2F%284%29+ and
x+=+%28%2B6.433%29%2F%284%29+
So
x+=-3.1 ---------------------discount negative lengths
and
x+=1.6 inches-----------------width
2x%2B3+=2%281.6%29%2B3=6.2inches----------- length
CK
A=10=(1.6)*(6.2)
~10=10


Hope this helps----ptaylor