The sum of all real values of x satisfying the equation = 1 is

Engineering

Mathematics

Exponential and Basic Maths questions

Question

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

– 4

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

(i) If x² – 5x + 5 = 1 ⇒ x² – 5x + 4 = 0

⇒ x = 1 or 4

(ii) If x² + 4x – 60 = 0 ⇒ (x + 10) (x – 6) = 0

⇒ x = – 10 or 6

(iii) If x² – 5x + 5 = – 1

and x² + 4x – 60 = even integer

⇒ x² – 5x + 6 =0

⇒ x = 2 or x = 3

For x = 2,

x² + 4x – 60 = 4 + 8 – 60 = – 48 = even

For x = 3,

x² + 4x – 60 = 9 + 12 – 60 = – 39 = odd

$∴$ x = 2 is also a solution.

Sum of values of x = 3.

Understanding the Problem

We are given the equation: ${(x^{2} - 5 x + 5)}^{x^{2} + 4 x - 60} = 1$ and need to find the sum of all real values of x that satisfy it.

An expression of the form $a^{b} = 1$ can be true in three distinct cases:

Case 1: The base $a = 1$ (because $1^{b} = 1$ for any exponent b).
Case 2: The base $a = - 1$ and the exponent $b$ is an even integer (because $(^{- 1) even} = 1$ ).
Case 3: The exponent $b = 0$ and the base $a \neq 0$ (because $a^{0} = 1$ for any non-zero a).

We will solve our equation by analyzing each of these cases separately.

Step-by-Step Solution

Step 1: Define the Base and Exponent

Let's define the base and the exponent from our equation to make the cases clearer.

Base, $a = x^{2} - 5 x + 5$

Exponent, $b = x^{2} + 4 x - 60$

Our equation is $a^{b} = 1$ .

Step 2: Solve Case 1 (Base = 1)

Set the base equal to 1 and solve for x.

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

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

Factor the quadratic equation:

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

Solutions: $x = 1$ and $x = 4$

We must check if these values are valid by ensuring the base is defined (which it is for these values).

Valid solutions from Case 1: x = 1, x = 4

Step 3: Solve Case 2 (Base = -1 and Exponent is Even)

First, set the base equal to -1.

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

$x^{2} - 5 x + 6 = 0$

Factor the quadratic equation:

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

Potential solutions: $x = 2$ and $x = 3$

Now, for each of these, we must check the second condition: the exponent must be an even integer.

For x = 2:
Exponent, $b = 2^{2} + 4 (2) - 60 = 4 + 8 - 60 = - 48$

-48 is an even integer. x = 2 is a valid solution.
For x = 3:
Exponent, $b = 3^{2} + 4 (3) - 60 = 9 + 12 - 60 = - 39$

-39 is an odd integer, not even. x = 3 is NOT a valid solution.

Valid solution from Case 2: x = 2

Step 4: Solve Case 3 (Exponent = 0 and Base ≠ 0)

First, set the exponent equal to 0.

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

Factor the quadratic equation:

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

Potential solutions: $x = - 10$ and $x = 6$

Now, for each of these, we must check the second condition: the base must not be zero.

For x = -10:
Base, $a = (^{- 10) 2} - 5 (- 10) + 5 = 100 + 50 + 5 = 155 \neq 0$

x = -10 is a valid solution.
For x = 6:
Base, $a = 6^{2} - 5 (6) + 5 = 36 - 30 + 5 = 11 \neq 0$

x = 6 is a valid solution.

Valid solutions from Case 3: x = -10, x = 6

Detailed Solution

Understanding the Problem

Step-by-Step Solution

Step 1: Define the Base and Exponent

Step 2: Solve Case 1 (Base = 1)

Step 3: Solve Case 2 (Base = -1 and Exponent is Even)

Step 4: Solve Case 3 (Exponent = 0 and Base ≠ 0)

Step 5: Compile All Valid

Detailed Solution

Understanding the Problem

Step-by-Step Solution

Step 1: Define the Base and Exponent

Step 2: Solve Case 1 (Base = 1)

Step 3: Solve Case 2 (Base = -1 and Exponent is Even)

Step 4: Solve Case 3 (Exponent = 0 and Base ≠ 0)

Step 5: Compile All Valid

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