SOLUTION: Please help me solve this equation: The length of a rectangle is 5 meters longer than the width. If the area is 22 square meters, find the rectangles dimensions. Round to the neare

Algebra ->  Rectangles -> SOLUTION: Please help me solve this equation: The length of a rectangle is 5 meters longer than the width. If the area is 22 square meters, find the rectangles dimensions. Round to the neare      Log On


   

Question 242446: Please help me solve this equation: The length of a rectangle is 5 meters longer than the width. If the area is 22 square meters, find the rectangles dimensions. Round to the nearest tenth of a meter.
I think the equation below is the right equation to solve the problem but I am having issues solving the equation.
X%5E2%2B5X=22
Found 2 solutions by JimboP1977, stanbon:
Answer by JimboP1977(311) About Me  (Show Source):
You can put this solution on YOUR website!
You have the correct equation.
To solve you could use the quadratic formula or completing the square. I tend to use the latter as I find it less fiddly but that is just me!
%28x%2B%285%2F2%29%29%5E2+-+25%2F4+=+22
%28x%2B%285%2F2%29%29%5E2+=+22%2B25%2F4
x+=+sqrt%28113%2F4%29-5%2F2
x = 2.8m (rounding to the nearest tenth) This is the width and I am sure you can find the length!

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
I think the equation below is the right equation to solve the problem but I am having issues solving the equation.
X%5E2%2B5X=22
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x^2 + 5x - 22 = 0
Use the quadratic formula:
x = [-5 +- sqrt(25 - 4*-22)}/2
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x = [-5 +- sqrt(113)]/2
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x = [-5 +- 10.63)]/2
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Positive solution:
x = (-5 + 10.63]/2
x = 5.63 meters (width)
x+5 = 10.63 meters (length)
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Cheers,
Stan H.