Answer: C - The number of real solutions is 2. Use x2 − y2 = (x − y)(x + y)x4 − 1 = 0(x2 )2 − 12 = 0(x2 − 1)(x2 + 1) = 0x2 − 1 = 0x2 = 1x = ±√1 = ±1x2 + 1 = 0x2 = − 1x = ±√( − 1) = ±iSo, there are two real solutions.

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What is the number of real solutions of x⁴ − 1 = 0?

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Answer: C - The number of real solutions is 2. Use x² − y² = (x − y)(x + y)
x⁴ − 1 = 0
(x² )² − 1² = 0
(x² − 1)(x² + 1) = 0
x² − 1 = 0
x² = 1
x = ±√1 = ±1
x² + 1 = 0
x² = − 1
x = ±√( − 1) = ±i

So, there are two real solutions.

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