The sum of all values of $x$ that satisfy the equation $\left(x^{2}-5 x+5\right)^{x^{2}+4 x-60}=1$ is
(A) -4
(B) 3
(C) 1
(D) 5
(E) 6
The sum of all values of $x$ that satisfy the equation $\left(x^{2}-5 x+5\right)^{x^{2}+4 x-60}=1$ is
(A) -4
(B) 3
(C) 1
(D) 5
(E) 6

## Solution

We consider the solution in three cases.

Case 1 It is possible for the base to be 1.

Therefore, $x^{2}-5 x+5=1$

$x^{2}-5 x+4=0$

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

Therefore $x=1$ or $x=4$.

Both these values are acceptable for $x$.

Case 2 It is possible that the exponent be 0.

Therefore, $x^{2}+4 x-60=0$

$$
(x+10)(x-6)=0
$$

$$
x=-10 \text { or } x=6
$$

Note: It is easy to verify that neither $x=-10$ nor $x=6$ is a zero of $x^{2}-5 x+5$, $x^{2}-5 x+5=0$, so that the indeterminate form $0^{\circ}$ does not occur.

Case 3 It is possible that the base is -1 and the exponent is even.

Therefore, $x^{2}-5 x+5=-1$ but $x^{2}+4 x-60$ must also be even.

$$
x^{2}-5 x+5=-1
$$

$$
\begin{aligned}
& x^{2}-5 x+6=0 \\
&(x-2)(x-3)=0 \\
& x=2 \text { or } x=3
\end{aligned}
$$

If $x=2$, then $x^{2}-4 x-60$ is even, so $x=2$ is a solution.

If $x=3$, then $x^{2}-4 x-60$ is odd, so $x=3$ is not a solution.

Therefore the sum of the solutions is $1+4-10+6+2=3$.

ANSWER: (B)