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

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

(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.