The sum of all real values of x satisfying th | vedclass

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(C) The equation is of the form $f(x)^{g(x)} = 1$. This holds true in three cases:Case $1$: The base is $1$. $x^2 - 5x + 5 = 1 \Rightarrow x^2 - 5x +

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$3$

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(C) The equation is of the form $f(x)^{g(x)} = 1$. This holds true in three cases:Case $1$: The base is $1$. $x^2 - 5x + 5 = 1 \Rightarrow x^2 - 5x + 4 = 0 \Rightarrow (x-1)(x-4) = 0$. Thus,$x = 1, 4$.Case $2$: The base is $-1$ and the exponent is an even integer. $x^2 - 5x + 5 = -1 \Rightarrow x^2 - 5x + 6 = 0 \Rightarrow (x-2)(x-3) = 0$. Thus,$x = 2, 3$. Checking the exponent $g(x) = x^2 + 4x - 60$: For $x=2$,$g(2) = 4 + 8 - 60 = -48$ (even). For $x=3$,$g(3) = 9 + 12 - 60 = -39$ (odd). So,$x=3$ is rejected.Case $3$: The exponent is $0$ and the base is non-zero. $x^2 + 4x - 60 = 0 \Rightarrow (x+10)(x-6) = 0$. Thus,$x = -10, 6$. Checking the base $f(x) = x^2 - 5x + 5$: For $x=-10$,$f(-10) = 100 + 50 + 5 = 155 
eq 0$. For $x=6$,$f(6) = 36 - 30 + 5 = 11 
eq 0$. Both are valid.The set of all real values of $x$ is ${1, 4, 2, -10, 6}$.The sum is $1 + 4 + 2 - 10 + 6 = 3$.

The sum of all real values of $x$ satisfying the equation $(x^2 - 5x + 5)^{x^2 + 4x - 60} = 1$ is:

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