Question 204189: 1) The sum of the squares of two consecutive integers is 613. What are the integers?
Let x be one integer. Therefore x +1 is the next integer.
x^2+(x+1)^2= 613
2)The length of a rectangle is four more than the width. If the area of the rectangle is 621 m2, find the width and length of the rectangle.
Let w be the width. Therefore w + 4 is the length.
W(w +4) = 621
Found 2 solutions by RAY100, Theo:
Answer by RAY100(1637)   (Show Source):
You can put this solution on YOUR website! Very good setup
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1) x^2 +(x+1)^2 =613
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x^2 +x^2 +2x+ 1 =613
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2x^2 +2x -612 =0
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x^2 +x -306=0
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(x+18)(x-17) =0
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x = 17, -18
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Integers,,17,18,,,,,,,check,,17^2 +18^2 =613,,,ok
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also -18,,-17,,,,,,,check, (-17)^2 +(-18)^2 = 613,,,ok
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2) w(w+4) =621
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w^2 +4w -621=0
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(w+27)(w-23) = 0
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w = 23, -27 (not realistic)
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Dim"s, w=23,,,L=27
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check , 23*27 = 621,,,ok
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Answer by Theo(13342)  (Show Source):
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