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Zorluk: Çok zorQuadratic Equations and Polynomial Factoring

If xx is a real number such that (x25x+5)x29x+20=1(x^2 - 5x + 5)^{x^2 - 9x + 20} = 1, what is the sum of all distinct real values of xx that satisfy the equation?

  1. A
    10
  2. B
    12
  3. C
    14
  4. 15Cevap
  5. E
    9

Cevap

The sum of all distinct real values of xx that satisfy the equation is 15.
The equation (x25x+5)x29x+20=1(x^2 - 5x + 5)^{x^2 - 9x + 20} = 1 holds under three distinct conditions: when the exponent is 0 and the base is non-zero (x=4,5x = 4, 5), when the base is 1 (x=1,4x = 1, 4), and when the base is 1-1 provided the exponent is an even integer (x=2,3x = 2, 3, yielding even exponents 6 and 2 respectively). The set of distinct solutions is {1,2,3,4,5}\{1, 2, 3, 4, 5\}, and their sum is 15.

Adım Adım Çözüm

1
Analyze the general algebraic conditions under which an expression of the form uv=1u^v = 1 holds for real numbers uu and vv.
The equation uv=1u^v = 1 is satisfied in three distinct cases: (1) v=0v = 0 and u0u \neq 0, (2) u=1u = 1, or (3) u=1u = -1 with vv being an even integer.
Any non-zero real number raised to the power of 0 equals 1, 1 raised to any power equals 1, and 1-1 raised to an even integer power equals 1.
2
Evaluate Case 1 where the exponent x29x+20=0x^2 - 9x + 20 = 0.
Factoring gives (x4)(x5)=0(x - 4)(x - 5) = 0, yielding x=4x = 4 and x=5x = 5. Testing the base x25x+5x^2 - 5x + 5 at x=4x = 4 gives 101 \neq 0, and at x=5x = 5 gives 505 \neq 0. Thus x=4x = 4 and x=5x = 5 are valid solutions.
Exponent equal to zero produces u0=1u^0 = 1 as long as the base u0u \neq 0.
3
Evaluate Case 2 where the base x25x+5=1x^2 - 5x + 5 = 1.
Subtracting 1 gives x25x+4=0x^2 - 5x + 4 = 0, which factors as (x1)(x4)=0(x - 1)(x - 4) = 0, yielding x=1x = 1 and x=4x = 4. Both are valid.
Base equal to 1 raised to any real exponent equals 1.
4
Evaluate Case 3 where the base x25x+5=1x^2 - 5x + 5 = -1 and check if the exponent is an even integer.
Adding 1 gives x25x+6=0x^2 - 5x + 6 = 0, which factors as (x2)(x3)=0(x - 2)(x - 3) = 0, yielding x=2x = 2 and x=3x = 3. Testing x=2x = 2 in the exponent gives 229(2)+20=62^2 - 9(2) + 20 = 6 (even). Testing x=3x = 3 gives 329(3)+20=23^2 - 9(3) + 20 = 2 (even). Thus x=2x = 2 and x=3x = 3 are also valid solutions.
Base 1-1 raised to an even integer power equals 1.
5
Combine all unique real values of xx and compute their sum.
The set of distinct real solutions is {1,2,3,4,5}\{1, 2, 3, 4, 5\}. Their sum is 1+2+3+4+5=151 + 2 + 3 + 4 + 5 = 15.
Summing all identified distinct valid roots.

Anahtar Kavram

Solving polynomial exponential equations of the form f(x)g(x)=1f(x)^{g(x)} = 1 by systematically testing base-exponent cases.
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