Solving the Equation with Exponents

We are asked to find the sum of all real values of \(x\) that satisfy the equation:

\(\rm (x^2 - 5x + 5) ^{x^2 + 4x - 60 }= 1\)

This equation is in the form \(a^b = 1\), where the base is \(a = x^2 - 5x + 5\) and the exponent is \(b = x^2 + 4x - 60\). An equation of the form \(a^b = 1\) holds true for real numbers under the following conditions:

• Case 1: The base \(a\) is equal to 1. That is, \(x^2 - 5x + 5 = 1\).
• Case 2: The exponent \(b\) is equal to 0, provided the base \(a\) is not equal to 0. That is, \(x^2 + 4x - 60 = 0\) and \(x^2 - 5x + 5 \neq 0\).
• Case 3: The base \(a\) is equal to -1, and the exponent \(b\) is an even integer. That is, \(x^2 - 5x + 5 = -1\) and \(x^2 + 4x - 60\) is even.

Let's analyze each case to find the real values of \(x\).

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

In this case, the base is 1, so any real exponent will result in 1 (except for potential issues with \(1^{\pm \infty}\), but the exponent here is a polynomial, always finite for finite \(x\)).

We solve the quadratic equation \(x^2 - 5x + 5 = 1\):

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

We can factor this quadratic equation:

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

This gives us two possible values for \(x\):

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

Let's check these solutions in the original equation:

• For \(x = 1\): Base = \(1^2 - 5(1) + 5 = 1\). Exponent = \(1^2 + 4(1) - 60 = 1 + 4 - 60 = -55\). Equation becomes \(1^{-55} = 1\), which is true. So \(x=1\) is a solution.
• For \(x = 4\): Base = \(4^2 - 5(4) + 5 = 16 - 20 + 5 = 1\). Exponent = \(4^2 + 4(4) - 60 = 16 + 16 - 60 = 32 - 60 = -28\). Equation becomes \(1^{-28} = 1\), which is true. So \(x=4\) is a solution.

From Case 1, we have the solutions \(x = 1\) and \(x = 4\).

Case 2: Exponent Equals Zero (\(x^2 + 4x - 60 = 0\) and \(x^2 - 5x + 5 \neq 0\))

In this case, if the exponent is 0, the expression \(a^0 = 1\), provided the base \(a\) is not zero (\(0^0\) is generally considered undefined or 1 depending on context, but for continuous functions leading to this form, it's often 1; however, the strict condition for \(a^b=1\) is \(a \neq 0\)).

We solve the quadratic equation \(x^2 + 4x - 60 = 0\):

We look for two numbers that multiply to -60 and add up to 4. These numbers are 10 and -6.

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

This gives us two possible values for \(x\):

• \(x + 10 = 0 \implies x = -10\)
• \(x - 6 = 0 \implies x = 6\)

Now we must check if the base \(x^2 - 5x + 5\) is non-zero for these values:

• For \(x = -10\): Base = \((-10)^2 - 5(-10) + 5 = 100 + 50 + 5 = 155\). Since \(155 \neq 0\), \(x = -10\) is a valid solution. The equation becomes \(155^0 = 1\), which is true.
• For \(x = 6\): Base = \(6^2 - 5(6) + 5 = 36 - 30 + 5 = 11\). Since \(11 \neq 0\), \(x = 6\) is a valid solution. The equation becomes \(11^0 = 1\), which is true.

From Case 2, we have the solutions \(x = -10\) and \(x = 6\).

Case 3: Base Equals Negative One (\(x^2 - 5x + 5 = -1\) and \(x^2 + 4x - 60\) is even)

In this case, if the base is -1, the expression \((-1)^b = 1\) only if the exponent \(b\) is an even integer.

We solve the quadratic equation \(x^2 - 5x + 5 = -1\):

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

We can factor this quadratic equation:

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

This gives us two possible values for \(x\):

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

Now we must check if the exponent \(x^2 + 4x - 60\) is an even integer for these values:

• For \(x = 2\): Exponent = \(2^2 + 4(2) - 60 = 4 + 8 - 60 = 12 - 60 = -48\). Since -48 is an even integer, \(x = 2\) is a valid solution. The equation becomes \((-1)^{-48} = 1\), which is true.
• For \(x = 3\): Exponent = \(3^2 + 4(3) - 60 = 9 + 12 - 60 = 21 - 60 = -39\). Since -39 is an odd integer, \(x = 3\) is NOT a valid solution in this case (\((-1)^{-39} = -1\)).

From Case 3, we have the solution \(x = 2\).

Identifying Valid Real Solutions

Combining the valid solutions from all three cases, the set of all real values of \(x\) satisfying the equation is \(\{1, 4, -10, 6, 2\}\).

Calculating the Sum of Solutions

The real values of \(x\) that satisfy the equation are \(1, 4, -10, 6,\) and \(2\).

The sum of these values is:

Sum = \(1 + 4 + (-10) + 6 + 2\)

Sum = \(5 - 10 + 6 + 2\)

Sum = \(-5 + 6 + 2\)

Sum = \(1 + 2\)

Sum = \(3\)

Revision Table: Key Concepts for Solving Exponent Equations

Additional Information: Properties of Exponents and Equations

Understanding exponent properties is crucial for solving such equations. Recall that for real numbers:

• Any non-zero number raised to the power of 0 is 1 (\(a^0 = 1\) for \(a \neq 0\)).
• 1 raised to any real power is 1 (\(1^b = 1\)).
• -1 raised to an even integer power is 1 (\((-1)^{2n} = 1\)).
• -1 raised to an odd integer power is -1 (\((-1)^{2n+1} = -1\)).

When solving equations involving variables in both the base and the exponent, it's important to consider all these cases systematically and verify the solutions obtained against the specific conditions of each case, especially concerning the base being zero or the exponent being an integer (even/odd) when the base is negative.

Solving quadratic equations, as seen in the steps above, is a fundamental skill needed for this type of problem. Factoring or using the quadratic formula (\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)) are common methods for solving \(ax^2 + bx + c = 0\).