SOLUTION: the length of a rectangle is 11 centimeters less than six times its width. Its area is 21 square centimeters. find the dimensions of the rectangle

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Question 357214: the length of a rectangle is 11 centimeters less than six times its width. Its area is 21 square centimeters. find the dimensions of the rectangle
Found 2 solutions by mananth, nerdybill:
Answer by mananth(16946) (Show Source):
You can put this solution on YOUR website!
let width be x
length = 6x-11
Area = 21
..
x(6x-11)=21
6x^2-11x=21
6x^2-11x-21=0
6x^2-18x+7x-21=0
6x(x-3)+7(x-3)=0
(x-3)(6x+7)=0
x= 3 cm the width
length = 6x-11
18-11
7 cm the length

Answer by nerdybill(7384) (Show Source):
You can put this solution on YOUR website!
the length of a rectangle is 11 centimeters less than six times its width. Its area is 21 square centimeters. find the dimensions of the rectangle
.
Let w = width
then because "length of a rectangle is 11 centimeters less than six times its width" we have
6w-11 = length
.
w(6w-11) = 21

6w^2-11w = 21
6w^2-11w-21 = 0
applying the quadratic formula we get:
w = {3, -1.167}
we can toss out the negative solution leaving
w = 3 centimeters (width)
.
Length:
6w-11 = 6(3)-11 = 18-11 = 7 centimeters
.
.
Details of quadratic follows:

Solved by pluggable solver: SOLVE quadratic equation with variable

Quadratic equation (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=625 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 3, -1.16666666666667. Here's your graph: