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

The equation $(x^2-5x+5{)}^{x^2+4x-60}=1$ is of the form $a^b=1$. To solve this, we must recognize that there are three distinct logical scenarios where this equality holds.

We must investigate each case with the rigor of a detective to ensure no solutions are missed and no invalid roots are accepted.

The equation $a^b=1$ is satisfied under the following conditions:

1. The exponent $b$ is zero, provided the base $a eq 0$. 2. The base $a$ is exactly $1$. 3. The base $a$ is $-1$, provided the exponent $b$ is an even integer.

The Exponent is Zero

We set the exponent to zero:

Factoring the quadratic, we look for numbers that multiply to $-60$ and add to $4$:

This yields $x=-10$ and $x=6$. We must verify that the base $a=x^2-5x+5$ is not zero for these values.

For $x=-10$, $a = (-10)^2 - 5(-10) + 5 = 100 + 50 + 5 = 155 eq 0$. For $x=6$, $a = (6)^2 - 5(6) + 5 = 36 - 30 + 5 = 11 eq 0$. Both values are valid.

The Base is One

We set the base to one:

Factoring the quadratic:

This yields $x=4$ and $x=1$. Since $1$ raised to any real power is $1$, both values are valid.

The Base is Negative One

We set the base to negative one:

Factoring the quadratic:

This yields candidates $x=2$ and $x=3$. We must now perform a parity check on the exponent $b=x^2+4x-60$.

For $x=2$: Since $-48$ is an even integer, $x=2$ is a valid solution.

For $x=3$: Since $-39$ is an odd integer, $x=3$ is invalid.

The set of valid roots is $\{-10,6,1,4,2\}$. The sum of these roots is:

The final answer is 3.