SOLUTION: Quadratic Equations 3. The length of a rectangle is 16cm greater than its width. The area is 35m^2. Find the dimensions of the rectangle, to the nearest hundredth of a metre. Tha

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: Quadratic Equations 3. The length of a rectangle is 16cm greater than its width. The area is 35m^2. Find the dimensions of the rectangle, to the nearest hundredth of a metre. Tha      Log On


   

Question 198327: Quadratic Equations
3. The length of a rectangle is 16cm greater than its width. The area is 35m^2. Find the dimensions of the rectangle, to the nearest hundredth of a metre.
Thanks
Found 2 solutions by nerdybill, RAY100:
Answer by nerdybill(7384) About Me  (Show Source):
You can put this solution on YOUR website!
3. The length of a rectangle is 16cm greater than its width. The area is 35m^2. Find the dimensions of the rectangle, to the nearest hundredth of a metre.
.
Let w = width
then
w+16 = length
.
w(w+16) = 35
w^2+16w = 35
w^2+16w-35 = 0
Using the quadratic equation, we get:
x = {1.95, -17.95}
We can toss out the negative solution leaving:
x = 1.95 m
.
Details of quadratic follows:
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B16x%2B-35+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%2816%29%5E2-4%2A1%2A-35=396.

Discriminant d=396 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-16%2B-sqrt%28+396+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%2816%29%2Bsqrt%28+396+%29%29%2F2%5C1+=+1.9498743710662
x%5B2%5D+=+%28-%2816%29-sqrt%28+396+%29%29%2F2%5C1+=+-17.9498743710662

Quadratic expression 1x%5E2%2B16x%2B-35 can be factored:
1x%5E2%2B16x%2B-35+=+1%28x-1.9498743710662%29%2A%28x--17.9498743710662%29
Again, the answer is: 1.9498743710662, -17.9498743710662. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B16%2Ax%2B-35+%29

Answer by RAY100(1637) About Me  (Show Source):
You can put this solution on YOUR website!
Area = Length * Width
.
A = ( w+16) * ( w)
.
35 = w^2 + 16w
.
0 = w^2 +16w -35
.
using quadratic equation
.
a = 1,,b = 16,,,c=-35
.
w = { - (16) +/- sq rt ( (16)^2 - 4 (1) (-35) } / 2 (1)
.
w = { -16 +/- 19.899 } /2
.
w = -17.949, + 1.949,,,,-17.95, +1.95 ,,,,,neg dim not reasonable therefore w= 1.95
.
L = w+16 = 17.95
.
checking
(-17.95) (-17.95 +16 ) = 35,,,ok
.
(1.95) ( 1.95 + 16 ) = 35,,,,ok
.
A = 1.95 * 17.95 = 35,,,,ok