Question:

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

Updated On: Feb 14, 2025
  • 3
  • -4
  • 6
  • 5
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The Correct Option is A

Solution and Explanation

$\left(x^{2} - 5x + 5\right)^{x^2 +4x-60} = 1 = \left(x^{2} - 5x +5 \right)^{0} $
$\Rightarrow x^{2} + 4x -60 = 0 \left[a^{x} = a^{y} \Rightarrow x=y \, \, \, if \, a \neq 1, 0 , -1\right] $
$x =- 10, 6$
& base $ x^{2} -5x +5 = 0 $ or $1$ or $-1$
If $x^2 - 5x + 5 = 0 $


But it will not satisfy original equation .
Hence solutions are - $10, 6, 4, 1, 2$
So, sum of solutions $= - 10 + 6 + 4 + 1 + 2 = 3$
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Concepts Used:

Complex Numbers and Quadratic Equations

Complex Number: Any number that is formed as a+ib is called a complex number. For example: 9+3i,7+8i are complex numbers. Here i = -1. With this we can say that i² = 1. So, for every equation which does not have a real solution we can use i = -1.

Quadratic equation: A polynomial that has two roots or is of the degree 2 is called a quadratic equation. The general form of a quadratic equation is y=ax²+bx+c. Here a≠0, b and c are the real numbers.