Inverse hyperbolic functions

[Identity]

Consider y = cosh^-1(x). Then x = cosh(y) = 1/2(e^y + e^-y).
Multiplying e^y gives: x(e^y) = (e^y + e^-y)e^y/2.
Simplified you get: (e^y)^2 - 2xe^y + 1 = 0.
I think this is right so far. Solving the quadratic expression in e^y I get: e^y = x + sqrt(2x^2 - 2). The text says e^y = x + sqrt(x^2 - 1).
What did I do wrong there?

[HallsofIvy]

Consider y = cosh^-1(x). Then x = cosh(y) = 1/2(e^y + e^-y).
Multiplying e^y gives: x(e^y) = (e^y + e^-y)e^y/2.
Simplified you get: (e^y)^2 - 2xe^y + 1 = 0.
I think this is right so far. Solving the quadratic expression in e^y I get: e^y = x + sqrt(2x^2 - 2). The text says e^y = x + sqrt(x^2 - 1).
What did I do wrong there?
I can't tell what you did wrong since you don't show HOW you solved the quadratic!
You have a quadratic u^2- 2xu + 1= 0. a= 1, b= -2x and c= 1: the quadratic formula gives u= (-(-2x)+/- sqrt((-2x)^2- 4(1(1))/2= (2x+/- sqrt(4x^2-4))/2= (2x+/- 2sqrt(x^2-1)/2= x +/- sqrt(x^2- 1). Since an exponential is positive, e^y= x+ sqrt(x^2- 1).

[Identity]

(2x+/- 2sqrt(x^2-1)/2= x +/- sqrt(x^2- 1). Should the left hand side of that equation have 2x^2-2 in the brackets? If not, I'm not sure I understand it.

[HallsofIvy]

No, it shouldn't! And, since, once again, you say nothing about WHY you think it should, I'm not at all clear how to answer you.

For quadratic equation ax^2+ bx+ c= 0, the quadratic formula gives
x= (-b+/- sqrt{b^2- 4ac})/(2a) as solution. Your equation is (e^y)^2 - 2xe^y + 1 = 0, which gives the quadratic equation u^2+ 2xu+ 1= 0. That is, a= 1, b= 2x, and c= 1 so u= (-2x+/- sqrt{4x^2- 4(1)(1)})(2)= (-2x+/- 2sqrt{x^2- 1})/2= -x+/- sqrt{x^2-1}.

[Identity]

How the heck do you go from (-2x+/- sqrt{4x^2- 4})/(2), to (-2x+/- 2sqrt{x^2- 1})/2 ?

[Identity]

Oh, you sqrt the 4s and then distribute.