Question

For equation \[{({x^2} - 5x + 5)^{{x^2} + 4x - 60}} = 1\]
A. Number of real value of \[x = 4\]
B. Number of real value of \[x = 5\]
C. Sum of real value of \[x = 1\]
D. Sum of real value of \[x = 3\]

Answer 1

Hint: We need to solve the given polynomial equation by using factorization and find the possible factor to get the real value of \[x.\] Factorization is the process of finding factors of a given value. Factors are the integers which are multiplied to produce an original number.

Complete step by step solution:
The given polynomial expression, \[{({x^2} - 5x + 5)^{{x^2} + 4x - 60}} = 1\]
Let the possible factors, when
 \[{x^2} - 5x + 5 \ne 0\] and \[{x^2} + 4x - 60 = 0\] (or)
 \[{x^2} - 5x + 5 = 1\] and \[{x^2} + 4x - 60 = 0\] (or)
 \[{x^2} - 5x + 5 = - 1\] and \[{x^2} + 4x - 60 = \] Even integer.
To solve the above cases to find the real value,
Case I : When \[{x^2} - 5x + 5 \ne 0\] and \[{x^2} + 4x - 60 = 0\]
 \[ \Rightarrow {x^2} + 4x - 60 = 0\]
By using factorization, we get
 \[
   \Rightarrow {x^2} + 10x - 6x - 10 = 0 \\
   \Rightarrow x(x + 10) - 6(x + 10) = 0 \;
 \]
By grouping the factor and get two values of \[x\] .
 \[ \Rightarrow (x + 10)(x - 6) = 0\]
Therefore the value is \[x = - 10,x = 6\]
 \[{x^2} - 5x + 5 \ne 0\]
Case II : When \[{x^2} - 5x + 5 = 1\]
 \[ \Rightarrow {x^2} - 5x + 5 = 1\]
To simplify the above expression,
 \[
   \Rightarrow {x^2} - 5x + 5 - 1 = 0 \\
   \Rightarrow {x^2} - 5x + 4 = 0 \;
 \]
By using factorization, we get
 \[
   \Rightarrow {x^2} - 4x - x + 4 = 0 \\
   \Rightarrow x(x - 4) - 1(x - 4) = 0 \;
 \]
By grouping the factor and get two values of \[x\] .
 \[ \Rightarrow (x - 4)(x - 1) = 0\]
Therefore the value is \[x = 4,x = 1\] .
Case III : When \[{x^2} - 5x + 5 = - 1\] , \[{x^2} + 4x - 60 = \] Even integer.
 \[ \Rightarrow {x^2} - 5x + 5 = - 1\]
To simplify the above expression,
 \[
   \Rightarrow {x^2} - 5x + 5 + 1 = 0 \\
   \Rightarrow {x^2} - 5x + 6 = 0 \;
 \]
By using factorization, we get
 \[
   \Rightarrow {x^2} - 2x - 3x + 6 = 0 \\
   \Rightarrow x(x - 2) - 3(x - 2) = 0 \;
 \]
By grouping the factor and get two values of \[x\] .
 \[ \Rightarrow (x - 3)(x - 2) = 0\]
Therefore the value is \[x = 3,x = 2\] .
Now, when \[x = 2\] substitute in \[{x^2} + 4x - 60 = \] even integer.
 \[{x^2} + 4x - 60 = {(2)^2} + 4(2) - 60 = 4 + 8 - 60 = - 48\] , which is an even integer.
when \[x = 3\] substitute in \[{x^2} + 4x - 60 = {(3)^2} + 4(3) - 60 = 9 + 12 - 60 = - 39\]
which is not an even integer.
So, In this case, we get \[x = 2\] .
Finally, The sum of all real values of \[x = - 10 + 6 + 4 + 1 + 2 = 3\] .
The final answer is option (D) Sum of real value of \[x = 3\] .
So, the correct answer is “Option D”.

Note: Here, we use factorization method of given polynomial expression by solving three possible cases and then apply the values into the given equation to get the final answer is sum of all real values.