The roots of the equation $(p - 2)x^2 + 2(p - 2)x + 2 = 0$ are not real when
- $p \in [1, 2]$
- $p \in [2, 3]$
- $p\in (2, 4)$
- $p \in [3, 4]$
Reveal answer
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C
Correct answer
Explanation
For non-real roots, the discriminant D < 0. D = [2(p-2)]^2 - 4(p-2)(2) = 4(p-2)^2 - 8(p-2) = 4(p-2)(p-2-2) = 4(p-2)(p-4). For D < 0, (p-2)(p-4) < 0, which occurs when 2 < p < 4.
AI explanation
The roots are not real when the discriminant is strictly less than zero, meaning b squared minus four a c is less than zero. Substituting the coefficients from the equation gives four times the quantity p minus two squared minus eight times p minus two is less than zero. Factoring yields four times p minus two times p minus four is less than zero, which solves to p is between two and four. Therefore, the roots are not real for p in the interval from two to four, exclusive.