SOLUTION: Solve the following application problem. Show and explain the variable equation used and each each step involved in solving the equation. The width of a rectangle is 4 cm less

Question 470950: Solve the following application problem. Show and explain the variable equation used and each each step involved in solving the equation.
The width of a rectangle is 4 cm less than its length. If the area of the rectange is 87 cm^2, find the dimensions of the rectangle to the nearest thousandth.
Found 2 solutions by jorel1380, lwsshak3:
Answer by jorel1380(3719) (Show Source): You can put this solution on YOUR website!
l(l-4)=87
l2-4l-87=0

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=364 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 11.5393920141695, -7.53939201416946. Here's your graph:

Throwing out the negative answer, we get the length of the rectangle to be 11.5393920141695 cm, and the width to be 7.5393920141695 cm..
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
Solve the following application problem. Show and explain the variable equation used and each each step involved in solving the equation.
The width of a rectangle is 4 cm less than its length. If the area of the rectange is 87 cm^2, find the dimensions of the rectangle to the nearest thousandth.
**
let x=length of rectangle
x-4 =width of rectangle
area of rectangle=length*width=87 cm ^2
x(x-4)=87
x^2-4x-87=0
solve by quadratic formula as follows:
..

..
a=1, b=-4, c=-87
x=[-(-4)�sqrt((-4)^2-(4*1*(-87)]/2*1
x=[4�sqrt(16+348)]/2
x=(4�√364)/2
x=(4�19.079)/2
x=11.539
x-4=7.539
or
x=-7.534 (reject, length>0)
ans:
length of rectangle=11.539 cm
width of rectangle=7.539 cm
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