SOLUTION: A right triangle has sides measuring 2x-1, 3x-13, and 3x-4 units. What is the value of x?

Question 378651: A right triangle has sides measuring 2x-1, 3x-13, and 3x-4 units. What is the value of x?
Found 2 solutions by stanbon, richard1234:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
A right triangle has sides measuring 2x-1, 3x-13, and 3x-4 units. What is the value of x?
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longest side: 3x-4
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Equation:
(2x-1)^2 + (3x-13)^2 = (3x-4)^2
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4x^2-4x+1 + 9x^2-78x+169 = 9x^2-24x+16
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13x^2-82x+170 = 9x^2-24x+16
4x^2 - 58x + 154 = 0
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2x^2 - 29x + 77 = 0
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Positive Solution:
x = [29 +- sqrt(29^2 - 4*2*77)]/4
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x = [29 +- sqrt(225)]/4
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x = [29+-15]/4
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x = 11 or x = 7/2
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Only x=11 is a viable answer to get positive side lengths.
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Cheers,
Stan H.

Answer by richard1234(7193) (Show Source): You can put this solution on YOUR website!
First we must find which side must be the hypotenuse. Clearly, 3x - 13 cannot be the hypotenuse as 3x - 4 is larger. Therefore either 2x - 1 or 3x - 4 is larger. Furthermore, since all the sides are positive, 3x - 13 > 0 --> x > 13/3. We can see that for all values of x larger than 13/3, 3x - 4 > 2x - 1, so 3x - 4 is the hypotenuse.
Now we can solve for x using the Pythagorean theorem. We have

x = 7/2 or x = 11
However, if x = 7/2, then 3x - 13 is negative -- therefore the only value for x is x = 11.
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