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

-4

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Solution

The correct option is D
3

$x^{2} - 5 x + 5 = 1 \Rightarrow x = 1, 4 o r x^{2} - 5 x + 5 = - 1 \Rightarrow x = 2, 3 o r x^{2} + 4 x - 60 = 0 \Rightarrow x = - 10, 6$

$∴ x = 3$

will be rejected as L.H.S. becomes –1

So, sum of value of x = 1 + 4 + 2 – 10 + 6 = 3

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