Questions on Geometry: Length, distance, coordinates, metric length answered by real tutors!

You can put this solution on YOUR website!
If the length of a rectangular field is 8 feet more than double of its width. Find the dimensions of the field if its area is 540 square feet?
.
Let x = width
then from "length of a rectangular field is 8 feet more than double of its width"
2x+8 = length
.
x(2x+8) = 540
2x^2+8x = 540
x^2+4x = 270
x^2+4x-270= 0
.
Since you can't factor you must use the quadratic equation.
This will result in:
x = {14.55, -18.55}
.
Since width can't be negative:
x = 14.55 feet
.
Length:
2x+8 = 2(14.55)+8 = 37.1 feet
.
Dimensions: 14.55 by 37.1 feet
.
Details of quadratic:

Solved by pluggable solver: SOLVE quadratic equation with variable

Quadratic equation ax^2+bx+c=0

(in our case 1x^2+4x+-270 = 0

) has the following solutons:

$x[12] = (b+-sqrt( b^2-4ac ))/2\a$

For these solutions to exist, the discriminant b^2-4ac

should not be a negative number.

First, we need to compute the discriminant b^2-4ac

.

Discriminant d=1096 is greater than zero. That means that there are two solutions: $x[12] = (-4+-sqrt( 1096 ))/2\a$ .

$x[1] = (-(4)+sqrt( 1096 ))/2\1 = 14.5529453572468$
$x[2] = (-(4)-sqrt( 1096 ))/2\1 = -18.5529453572468$

Quadratic expression 1x^2+4x+-270

can be factored:
1x+4x+-270 = 1(x-14.5529453572468)*(x--18.5529453572468)

Again, the answer is: 14.5529453572468, -18.5529453572468. Here's your graph:
graph( 500, 500, -10, 10, -20, 20, 1*x^2+4*x+-270 )