What is the number of roots of the equation (x^2-5x+5)^(x^2-5x-6)=1?

\(a^{b}=1\) under the following three conditions -

1) a = 1; b can be any value
2) b = 0; a can be any value except 0
3) a = -1; b has to be an even integer

Let's check all of these conditions for the above equation

\((x^2-5x+5)^{(x^2-5x-6)}=1\)

Case 1: \(x^2-5x+5 = 1\)

\(x^2 - 5x + 4 = 0\)

\(x^2 - 4x - x + 4 = 0\)

\((x-1)(x-4) = 0\)

• x = 1
  
• x = 4

Case 2: \(x^2-5x-6 = 0 \) AND \(x^2-5x+5 \neq 0\)

\(x^2-6x+x-6 = 0 \)

\((x+1)(x-6) = 0 \)

x = -1
Let's verify whether \(x^2-5x+5=0\)
\(1+5+5 = 11\)
\(11 \) → As the value is not equal to zero, x = -1 is a valid value.

x = 6
Let's verify whether \(x^2-5x+ 5 = 0\)
\(36-30+5\)
\(11 \) → As the value is not equal to zero, x = 6 is a valid value.

Case 3: \(x^2-5x+5 = -1 \) AND \(x^2-5x-6 = \text{even}\)

\(x^2-5x+6=0\)

\(x^2-3x-2x+6=0\)

\((x-3)(x-2)=0\)

x = 3 Let's verify whether \(x^2-5x-6 = \text{even}\)
odd - odd - even = even → As \(x^2-5x-6\) is even, x = 3 is a valid value.

x = 2 Let's verify whether \(x^2-5x-6 = \text{even}\)
even - even - even = even → As \(x^2-5x-6\) is even, x = 2 is a valid value.

Hence, a total of 6 values are possible.

Option E