How many distinct real solutions does the given equation have? x^2 = -841...

1. TRANSLATE the problem information

• Given equation: \(\mathrm{x^2 = -841}\)
• Need to find: How many distinct real solutions exist

2. INFER the key mathematical constraint

• Think about what \(\mathrm{x^2}\) means for any real number x
• Key insight: The square of any real number is always non-negative
• This means \(\mathrm{x^2 \geq 0}\) for every possible real value of x

3. APPLY CONSTRAINTS to determine possibility

• The equation requires \(\mathrm{x^2 = -841}\)
• But \(\mathrm{x^2}\) must be \(\mathrm{\geq 0}\) (from step 2)
• Since \(\mathrm{-841 \lt 0}\), this creates an impossible situation
• No real number x can satisfy this equation

Answer: D (Zero distinct real solutions)

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students may attempt to solve by taking the square root of both sides, writing \(\mathrm{x = \pm\sqrt{-841}}\), without recognizing that square roots of negative numbers don't exist in the real number system.

This misconception leads them to think there should be two solutions (one positive, one negative), causing them to select Choice B (Exactly two).

Second Most Common Error:

Conceptual confusion about number systems: Students might confuse real numbers with complex numbers, knowing that \(\mathrm{\sqrt{-841} = \pm 29i}\) in the complex system, but failing to recognize the problem specifically asks for real solutions.

This causes them to get stuck between thinking there are solutions versus no solutions, leading to guessing.

The Bottom Line:

This problem tests the fundamental understanding that squares of real numbers cannot be negative - a property so basic that students often overlook it when faced with an equation that seems "solvable" at first glance.