Determine the sum of all real numbers satisfying:
\((x^2-4x+2)^{x^2-5x+2}\)
This will happen when
x^2-5x+2 = 0 ( so long as x^2-4x+2 isn't 0 as well. )
use the quadratic formula
Ok, the sum will equal 9 in that case. But I think there are other cases too, for example when x^2-4x+2 = 1. Can you help me think of other ones?
\(x^2-5x+2 = 0\\ x = {-b \pm \sqrt{b^2-4ac} \over 2a}\\ x = {5 \pm \sqrt{25-8} \over 2}\\ \)
The sum of these is 5
Just checking that those answsers are valid. what can x not be
\(x^2-4x+2\ne0\\ x \ne {-b \pm \sqrt{b^2-4ac} \over 2a}\\ x \ne {4 \pm \sqrt{16-8} \over 2}\\\)
Yes the answer is fine.