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

Algebra ->  Pythagorean-theorem -> SOLUTION: The length of a rectangle exceeds twice its width by 3 inces. If the area is 10 square inches, find the rectangles dimensions. Round to the nearest tenth of an inch.      Log On


   

Question 108056This question is from textbook Introductoru & Intermediate Algebra
: The length of a rectangle exceeds twice its width by 3 inces. If the area is 10 square inches, find the rectangles dimensions. Round to the nearest tenth of an inch. This question is from textbook Introductoru & Intermediate Algebra

Found 2 solutions by solver91311, checkley71:
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!
Start with the standard formula for the area of a rectangle: A=lw

But we are told that the length is equal to twice the width plus 3 inches,

Therefore, l=2w%2B3

Since the area = 10 square inches, we can now write:

%282w%2B3%29w=10

Distribute and put the equation in standard form:

2w%5E2%2B3w-10=0

There is no neat and tidy way to factor this one, so we fall back on the good old quadratic formula:

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

Answer by checkley71(8403) About Me  (Show Source):
You can put this solution on YOUR website!
L=2W+3
L*W=10
(2W+3)*W=10
2W*2+3W-10=0
USING THE QUADRATIC EQUATON WE GET
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
X=(-3+-SQRT[3^2-4*2*-10])/2*2
X=(-3+-SQRT[9+80])/4
X=(-3+-SQRT89)/4
X=(-3+-9.43)/4
X=(-3+9.43)/4
X=6.43/4
X=1.6 ANSWER FOR THE WIDTH.
L=2*1.6
L=3.2+3
L=6.2 FOR THE LENGTH.
PROOF
1.6*6.2=10
10=10