SOLUTION: Paul and Joan wish to buy floor covering that covers 150 square feet for their large recreation room. They wish to cover a rectangle that is 5 feet longer than it is wide. How wide

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Question 1190473: Paul and Joan wish to buy floor covering that covers 150 square feet for their large recreation room. They wish to cover a rectangle that is 5 feet longer than it is wide. How wide should the rectangle be?
Found 2 solutions by ikleyn, math_tutor2020:
Answer by ikleyn(52781) About Me  (Show Source):
You can put this solution on YOUR website!
.
Paul and Joan wish to buy floor covering that covers 150 square feet for their large recreation room.
They wish to cover a rectangle that is 5 feet longer than it is wide. How wide should the rectangle be?
~~~~~~~~~~~~~~ 


This problem is to solve this quadratic equation  x*(x+5) = 150.


Or to find two integer numbers with the difference of 5 and the product of 150.


It can be guessed mentally in your head without using any equation/equations in 6 seconds 

(much faster than I write these lines).


The dimensions are 10 and 15 feet.    ANSWER

Solved.

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read the subject from these lessons
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    - PROOF of quadratic formula by completing the square
    - HOW TO solve quadratic equation by completing the square - Learning by examples
    - Solving quadratic equations without quadratic formula
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The referred lessons are the part of this textbook under the topic  "Quadratic equations".


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Answer by math_tutor2020(3817) About Me  (Show Source):
You can put this solution on YOUR website!

x = width
x+5 = length

area = length*width
area = x(x+5)
area = x^2+5x

set this equal to 150 and solve for x
x^2+5x = 150
x^2+5x-150 = 0

From here, you can use the quadratic formula with
a = 1, b = 5, c = -150

x+=+%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F%282a%29

x+=+%28-5%2B-sqrt%28%285%29%5E2-4%281%29%28-150%29%29%29%2F%282%281%29%29

x+=+%28-5%2B-sqrt%2825%2B600%29%29%2F%282%29

x+=+%28-5%2B-sqrt%28625%29%29%2F%282%29

x+=+%28-5%2Bsqrt%28625%29%29%2F%282%29 or x+=+%28-5-sqrt%28625%29%29%2F%282%29

x+=+%28-5%2B25%29%2F%282%29 or x+=+%28-5-25%29%2F%282%29

x+=+%2820%29%2F%282%29 or x+=+%28-30%29%2F%282%29

x+=+10 or x+=+-15

Ignore the negative x value because we cannot have a negative width.

x = 10 is the only practical solution

x = 10 = width
x+5 = 10+5 = 15 = length

area = length*width = 15*10 = 150
The answers are confirmed.