SOLUTION: 1.find two numbers whose sum is 13 and whose product is 40 2.the area of a rectangle is 94sq. units and its perimeter is 38units. find the length and width of the rectangle.

Question 874981: 1.find two numbers whose sum is 13 and whose product is 40
2.the area of a rectangle is 94sq. units and its perimeter is 38units.
find the length and width of the rectangle.

Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
1.find two numbers whose sum is 13 and whose product is 40
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Try pairs of factors of 40.
1*40 NG
2*20 NG
etc
====================
2.the area of a rectangle is 94sq. units and its perimeter is 38units.
find the length and width of the rectangle.
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P = 2L + 2W = 38
L + W = 19
L*W = 94
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L = 19 - W
Sub for L
(19-W)*W = 94
W^2 - 19W + 94 = 0

Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)

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 : .

The discriminant -15 is less than zero. That means that there are no solutions among real numbers.

If you are a student of advanced school algebra and are aware about imaginary numbers, read on.

In the field of imaginary numbers, the square root of -15 is + or - .

The solution is , or
Here's your graph:

============
No real number solution.
The max area of a rectangle of given perimeter is a square.
For P = 38, side length = 9.5
Area = 9.5^2 = 90.25 sq units
94 is not possible.
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