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

3

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- 4

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6

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5

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Solution

The correct option is A $3$
$(x^{2} - 5 x + 5)^{x^{2} + 4 x - 60} = 1$

First possibility is $x^{2} + 4 x - 60 = 0$
$x = - 10, 6$

Seond possibility is $x^{2} - 5 x + 5 = 1$
$x = 1, 4$

Third Possibility
$x^{2} - 5 x + 5 = - 1$
$x = 2, 3$

substitute $x = 3$ it will give odd number in $x^{2} + 4 x - 60$

So $x = 2$ is only acceptable.

Sum of the integers $= - 10 + 4 + 1 + 6 + 2 = 3$

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