SOLUTION: So I have a 30-60-90 triangle with the short leg that has a length of 5x and the long leg with a length of x^2+2x. I have to solve for x. (Hint) use what you know about the ratio.

Question 827964: So I have a 30-60-90 triangle with the short leg that has a length of 5x and the long leg with a length of x^2+2x. I have to solve for x. (Hint) use what you know about the ratio.
Answer by LinnW(1048) (Show Source): You can put this solution on YOUR website!
If short leg is 5x, hyp is twice or 10x
So we have ( x^2+2x )^2 + (5x)^2 = (10x)^2
x^4 +4x^3 + 4x^2 +25x^2 = 100x^2
subtract 100x^2 from each side
x^4 +4x^3 + 4x^2 +25x^2 - 100x^2 = 0
x^4 +4x^3 - 71x^2 = 0
Factor
x^2(x^2 + 4x -71)=0
Since x^2 is a factor, one solution is zero
Now solve x^2 + 4x -71 = 0
See http://www.wolframalpha.com/input/?i=x^2%2B4x-71%3D0

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