# Real VAlues

• Mar 1st 2007, 12:05 PM
Rimas
Real VAlues
Find all real values of x that satisfy (x^2-5x+5)^x^2-9x+2=1
• Mar 1st 2007, 01:00 PM
Jhevon
Quote:

Originally Posted by Rimas
Find all real values of x that satisfy (x^2-5x+5)^x^2-9x+2=1

Okay, we know that (or i hope we know that) anything raised to the zero gives us one. So that means our power x^2 - 9x + 2 = 0. Now we just solve the quadratic equation.

x^2 - 9x + 2 = 0
Using the quadratic formula:
=> x = [9 +/- sqrt{9^2 - 4(1)(2)}]/2
= [9 +/- sqrt(73)]/2
= 8.772 or 0.228

Plugging these into your calculator won't give you exactly zero, since they are decimal approximations, but you get -0.000016 or something like that, which is good enough. If you want an answer closer to zero, use more decimal points. We also want to double check these values for x into the base, that is into x^2 - 5x + 5, to see if they cause that to be zero. Since 0^0 is meaningless (i think), we would disregard any solution that is also a root of the base, both these solutions are ok.