The current state of math education (vanity)

Yahoo answers ^ | 10/28/2010 | self

[Catphish]

I’m certain x is restricted to real numbers here. Unless this was an undergrad taking complex variables and in that case he would probably use z not x. You are talking about something else in terms of the range of the function once you allow x to be complex.

[Bob]

You just multiply through by x, so x + 2/x = y becomes x^2+2=yx. Then you get -x^2+yx-2=0. The discriminant of that is y^2-8. Since you are looking for real values that y can be y^2-8 has to be greater than or equal to 0, i.e., the absolute value of y has to be greater than or equal to the square root of 8.

Thanks for the explanation. I think that the 'y' term in there was what confused me. Duh!

[PeteCat]

for the first the range is all real numbers except between -sqrt8 and sqrt8.
try r - real number candidate
r=x+(1/x)
multiply by x
rx=x^2 + 1
0 = x^2 -rx +1
x=(r+- sqrt(r^2 - 8))/2

using the quadratic formula
http://en.wikipedia.org/wiki/Quadratic_equation

the term under the sqrt has to be positive for real values.

[Grams A]

So we have a generation of High School students who cannot perform simple, basic mathematics without a calculator.”

Made a puchase recently of $1.72. Originally gave teen clerk $2.00 which she entered into her register which registered $.28 coming back to me. Then I found two pennies and gave her that assuming she would just give me $.30 back. She absolutely freaked out because she didn’t know what to do and had to call manager over to correct register. He made a big deal about it and stated, “The register must record the transaction exactly as it occurred” and gave me back $.28. So he - probably early 30’s - apparently didn’t know how to figure it out either. Unfortunately I am old enough to well remember when clerks had to calculate your change in their head and count it back to you. No wonder many stores are going to self check-out and credit cards only.