SOLUTION: The length of a rectangle is one less than twice the width. The area of the rectangle is 66. What are the dimensions of the rectangle? Only an algebraic solution will be accepted.

Algebra ->  Equations -> SOLUTION: The length of a rectangle is one less than twice the width. The area of the rectangle is 66. What are the dimensions of the rectangle? Only an algebraic solution will be accepted.      Log On


   

Question 1117918: The length of a rectangle is one less than twice the width. The area of the rectangle is 66. What are the dimensions of the rectangle? Only an algebraic solution will be accepted.
Found 2 solutions by josgarithmetic, Shin123:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
%282x-1%29x=66 for length 2x-1 and with x.

2x%5E2-x-66=0

x=%281%2B-sqrt%281%2B4%2A2%2A66%29%29%2F4
x=%281%2Bsqrt%281%2B8%2A66%29%29%2F4
x=%281%2Bsqrt%28529%29%29%2F4
x=%281%2B23%29%2F4
highlight%28x=6%29--------width
highlight%2811%29-------length


You could also find the same result by factoring.

Answer by Shin123(626) About Me  (Show Source):
You can put this solution on YOUR website!
So %282x-1%29%2Ax=66 2x-1 is the length and x is the width. Turn that to a quadratic equation,
2x%5E2-x-66=0 Use the quadratic formula.
x=%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F2a
x=%28-%28-1%29%2B-sqrt%28-1%5E2-4%2A2%2A-66%29%29%2F2%282%29
x=%281%2B-sqrt%281-8%2A-66%29%29%2F4
x=%281%2B-sqrt%28529%29%29%2F4
x=%281%2B-+23%29%2F4 Ignore the minus because we're looking for positive answers.
highlight%28x=6%29 The width is 66%2F6=highlight%2811%29