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

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Solution

The correct option is A 3
Given, $(x^{2} - 5 x + 5)^{x^{2} + 4 x - 60} = 1$
Clearly, this is possible when
$I . x^{2} + 4 x - 60 = 0 a n d x^{2} - 5 x + 5 \neq 0$
or
$I I . x^{2} - 5 x + 5 = 1$
or
$I I I . x^{2} - 5 x + 5 = - 1$ and $x^{2} + 4 x - 60$ = Even integer.
Case I: When $x^{2} + 4 x - 60 = 0$
$\Rightarrow x^{2} + 10 x - 6 x - 60 = 0$
$\Rightarrow x (x + 10) - 6 (x + 10) = 0 \Rightarrow (x + 10) (x - 6) = 0$
$\Rightarrow$ x=-10 or x =6
Note that, for these two values of $x, x^{2} - 5 x + 5 \neq 0$
Case II: When $x^{2} - 5 x + 5 = 1$
$\Rightarrow x^{2} - 5 x + 4 = 0$
$\Rightarrow x^{2} - 4 x - x + 4 = 0$
$\Rightarrow x (x - 4) - 1 (x - 4) = 0$
$\Rightarrow (x - 4) (x - 1) = 0 \Rightarrow x = 4$ or x=1
Case III: When $x^{2} - 5 x + 5 = - 1$
$\Rightarrow x^{2} - 5 x + 6 = 0 \Rightarrow x^{2} - 2 x - 3 x + 6 = 0$
$\Rightarrow$ x(x-2)-3(x-2)= 0
$\Rightarrow$ (x-2)(x-3)=0
$\Rightarrow$ x=2 or x=3
Now, when x = 2, $x^{2} + 4 x - 60 = 4 + 8 - 60 = - 48,$ which is an even integer.
When x = 3, $x^{2} + 4 x - 60 = 9 + 12 - 60 = - 39,$ which is not an even integer.
Thus, in this case, we get x = 2.
Hence, the sum of all real values of x = - 10 + 6 + 4 + 1 + 2 =3

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