# 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

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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.
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Found 2 solutions by nerdybill, RAY100:
Answer by nerdybill(7090)   (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 (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=396 is greater than zero. That means that there are two solutions: . Quadratic expression can be factored: Again, the answer is: 1.9498743710662, -17.9498743710662. Here's your graph:

Answer by RAY100(1637)   (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