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.

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) (Show Source): You can put this solution on YOUR website!
Start with the standard formula for the area of a rectangle:

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

Therefore,

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

Distribute and put the equation in standard form:

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

Answer by checkley71(8403) (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=(-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

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