# algebra

The hypotenuse of a right triangle is 8m long. One leg is 2m longer then the other. Find the lengths of the legs. How would you solve this equation?

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1. short leg --- x
longer leg --- x+2

x^2 + (x+2)^2 = 8^2
x^2 + x^2 + 4x + 4 - 64 = 0
2x^2 + 4x - 60 = 0
x^2 + 2x - 30 = 0
let's complete the square:
x^2 + 2x + 1 = 30+1 = 31
(x+1)^2 = 31
x+1 = ± √31
x = -1 ± √31
but x has to be positive,
so x = √31 - 1 = appr 4.568

so one side is 4.568, the other is 6.568

check:
4.568^2 + 6.568^2 = 64.0052
close enough to 64

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posted by Reiny

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