Question 176382
The width of a rectangle is 2 meters larger than its height. The diagonal brace measures sqrt 6m. Find the width and height. 
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Let height be "x".
Then width is "x+2"
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Pythagoras says x^2 + (x+2)^2 = (sqrt(6))^2
x^2 + x^2 + 4x + 4 = 6
2x^2 + 4x -2 = 0
x^2 + 2x -1 = 0
x = [-2 +- sqrt(4 - 4*1*-1)]/2
x = [-2 +- sqrt(8)]/2
x = [-1 +- sqrt(2)]
Positive solution:
x = -1 + 1.414 = 0.414 meters (height)
x+2 = 2.414 meters (width)
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Cheers,
Stan H.