Question 717127
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Do exactly what your teacher told you to do.  If the hypotenuse is *[tex \LARGE 2x\ -\ 2], then we can say *[tex \LARGE 2x\ -\ 2\ =\ c]. The designation as to the other two sides is arbitrary because of the commutativity of addition, so we can assign *[tex \LARGE x\ +\ 1\ =\ a] and *[tex \LARGE x\ =\ b].  Now plug these values into the Pythagorean relationship:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (2x\ -\ 2)^2\ =\ (x\ +\ 1)^2\ +\ x^2]


Now all you have to do is expand the squared binomials and collect like terms in the LHS.  That will leave you with a quadratic expression equal to zero, which is simply a quadratic equation.  Given correct arithmetic you will have an easily factorable quadratic equation.  Discard the zero root because that is clearly absurd, leaving you with a positive integer value for *[tex \LARGE x] that you can use to calculate the measures of your three triangle sides.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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