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

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

Clearly, this is possible when

I. $x^{2} + 4x - 60 = 0$ and $x^{2} - 5x + 5 \neq 0$

or

II. $x^{2} - 5x + 5 = 1$

III. $x^{2} - 5x + 5 = - 1$ and $x^{2} + 4x - 60$ = Even integer.

Case I  When $x^{2} + 4x - 60 = 0$

⇒                    $x^{2} + 10x - 6x - 60 = 0$

⇒                   $x(x + 10) - 6(x + 10) = 0$

⇒                   (x + 10)(x - 6) = 0

⇒                   x = -10 or x = 6

Note that, for these two values of $x, x^{2} - 5x + 5 \neq 0$

Case II  When  $x^{2} - 5x + 5 = 1$

⇒                      $x^{2} - 5x + 4 = 0$

⇒                      $x^{2} - 4x - x + 4 = 0$

⇒                      x(x - 4) -1(x - 4) = 0

⇒                      (x - 4)(x - 1) = 0 ⇒ x = 4 or x = 1

Case III  When   $x^{2} - 5x + 5 = -1$

⇒                        $x^{2} - 5x + 6 = 0$

⇒                        $x^{2} - 2x - 3x + 6 = 0$

⇒                        x(x - 2) - 3(x - 2) = 0

⇒                        (x - 2)(x - 3) = 0

⇒                        x = 2 or x = 3.

Now, when x = 2, $x^{2} + 4x - 60 = 4 + 8 - 60 = - 48$, which is an even integer.

When x = 3, $x^{2} + 4x - 60 = 9 + 12 - 60 = -39$, which is not an even integer.

Hence the sum of all real values of x = -10 + 6 + 4 + 1 + 2 = 3.

Last updated date: 23rd Sep 2023
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