does not exist
exist and is equal to 1
exist and is equal to -1/2
exist and is equal to 1/2
Given equation is 2x³ - x² + kx - 15 = 0 where k ∈ R
Since, any odd degree equation with real coefficients has at least one real root. So, real root exist.
Let α be the real root.
Since the coefficients are real, the complex roots occur in conjugate pairs. Therefore, if 2 + 3i is a root, then 2 - 3i is also a root.
Let the roots be α, 2 + 3i, 2 - 3i.
Sum of roots = α + (2 + 3i) + (2 - 3i) = α + 4
Product of roots = α(2 + 3i)(2 - 3i) = α(4 - 9i²) = α(4 + 9) = 13α
From the given equation, sum of roots = 1/2 and product of roots = 15/2
Therefore, α + 4 = 1/2 => α = 1/2 - 4 = -7/2
13α = 15/2 => α = 15/26
However, since the sum of roots and product of roots must be consistent, there's a discrepancy.
Let's use Vieta's formulas:
Sum of roots = α + 2 + 3i + 2 - 3i = α + 4 = 1/2
α = -7/2
Product of roots = α(2 + 3i)(2 - 3i) = α(13) = 15/2
α = 15/26
There is a contradiction. Let's check the sum of roots:
Sum of roots = -b/a = 1/2
Product of roots = -d/a = 15/2
Let the roots be α, β, γ
α + β + γ = 1/2
αβ + βγ + αγ = k/2
αβγ = 15/2
Since 2 + 3i is a root, 2 - 3i is also a root. Let α be the real root.
Then α + 4 = 1/2 which gives α = -7/2
Also, α(13) = 15/2, so α = 15/26. There is a contradiction. Let's solve it differently.
Let the roots be 2+3i, 2-3i, and r. Then by Vieta's formulas:
(2+3i) + (2-3i) + r = 1/2
4 + r = 1/2
r = -7/2
Thus, the real root is -7/2.