Determine the number of integer values for x that fulfill the equation $(x^2 - 4x + 6)^{x + 2} = 1$ .

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Official Explanation

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To find the integer solutions for the equation

(x^2 - 4x + 6)^{x + 2} = 1

, we analyze the base and the exponent. The expression can equal 1 in several scenarios: when the base is 1, when the base is -1 and the exponent is even, or when the base is 0 and the exponent is positive. First, we set the base equal to 1:

x^2 - 4x + 6 = 1

, which simplifies to

x^2 - 4x + 5 = 0

. The discriminant is

(-4)^2 - 4(1)(5) = 16 - 20 = -4

, indicating no real solutions. Next, we set the base to -1:…

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Comments

ClelioMay 2, 2025

Thank you very much!

ShivangiApr 14, 2025

This is definitely a test-day trap

ShehranMar 22, 2025

Thank you very much!

What this question tests

What it tests

Your grasp of exponent rules — multiplying and dividing powers, power-of-a-power, and negative/zero exponents.

Common trap

Adding exponents when multiplying bases that have the same exponent, or confusing (x^a)^b with x^(a·b).

Determine the number of integer values for x that fulfill the equation (x2−4x+6)x+2=1(x^2 - 4x + 6)^{x + 2} = 1(x2−4x+6)x+2=1.