SOLUTION: The length of a rectangle is 6 ft. Greater than its width.the area is 832 sq/ft. Find its dimensions

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Question 931655: The length of a rectangle is 6 ft. Greater than its width.the area is 832 sq/ft. Find its dimensions
Found 2 solutions by Alan3354, Stitch:
Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website!
The length of a rectangle is 6 ft. Greater than its width.the area is 832 sq/ft. Find its dimensions
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Try pairs of factors of 832.
Find a pair that differ by 6
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1*832 NG
2*416 NG
etc

Answer by Stitch(470) (Show Source):
You can put this solution on YOUR website!
Equation 1: The area of a rectangle
Equation 2:
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We are given A = 832.
Since equation 2 is already solved for L, plug (W+6) into equation 1 for L
Equation 1:

Multiply the W through.

Set the equation equal to zero by subtracting 832 from both sides.

Now we can use the quadratic equation

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=3364 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 26, -32. Here's your graph:

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Since we can only have a positive distance, 26ft is W.

Now plug 26 into equation 2 for W
Equation 2: