I have a right triangle with two sides: x-13 and x-26 along with a hypotenuse x
What is x? I get 13 and 65. Why does this feel un perfect.
What is the length of the sides?
I have a right triangle with two sides: x-13 and x-26 along with a hypotenuse x
What is x? I get 13 and 65. Why does this feel un perfect.
What is the length of the sides?
$\displaystyle (x-13)^2 + (x-26)^2 = x^2$
$\displaystyle (x^2-26x+169) + (x^2-52x+676) = x^2$
$\displaystyle 2x^2-78x+845)= x^2$
$\displaystyle x^2-78x+845= 0$
$\displaystyle (x-65)(x-13)= 0$
13 is not a possible solution because that would give us a side of length zero and a side of negative length. Therefore, the length of the hypotenuse, i.e., $\displaystyle x$, is 65.
Thank you for your replies. I just feel that it's such an unrefined way.. Getting two solutions and going with one because the other one doesn't work in a LOGICAL world..I'm not sure if you understand what I'm getting at...Is this really the only way?
I do understand the question and I knew that I should go with 65 but I don't understand what happens with my other solution?
Why do I have two solutions and toss one. In physics, when you speak of negative velocity, you are going in the other direction. Isn't there some sort of rule for that in math as well?
I understand these questions are more of philosophical nature than pure mathematical so feel free to ignore them unless you have an answer or are intrigued yourself. Thank you.
You have two solutions because indeed if x = 13 then (x-13)^2 + (x-26)^2 = x^2. The issue is whether this makes sense when appied specifically to the physical reality of a triangle. Implied in the original question is a restriction that the lengths of the sides must be greater than 0, and this is what causes you to throw out one set of answers. Is there any other way? No, not really - the algebra technique does not automatically throw out values of x where x-13 and/or x-26 are not greater than 0.