For the equation , , if one of the root is square of the other, then is equal to
- A.
- B.
- C.
- D.
Show answer and solution
Correct option: C
Solution
Let the roots be α and α². By Vieta's formulas:
Product of roots: α · α² = α³ = 3/3 = 1, so α³ = 1, giving α = 1 or the complex cube roots of unity (ω, ω²).
Since real root α = 1 gives sum = 1 + 1 = 2, then p = −3·2 = −6 < 0, which violates p > 0.
So take α = ω (primitive cube root of unity). Sum of roots = ω + ω² = −1, and p = −3·(−1) = 3.
Therefore p = 3.
Product of roots: α · α² = α³ = 3/3 = 1, so α³ = 1, giving α = 1 or the complex cube roots of unity (ω, ω²).
Since real root α = 1 gives sum = 1 + 1 = 2, then p = −3·2 = −6 < 0, which violates p > 0.
So take α = ω (primitive cube root of unity). Sum of roots = ω + ω² = −1, and p = −3·(−1) = 3.
Therefore p = 3.
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