If $z^{3} + (3 + 2 i) z + (- 1 + i a) = 0$ has one real root, then the value of $^{'} a^{'}$ lies in the interval
$(a \in R)$

(- 2, 1)

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(- 1, 0)

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(0, 1)

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(- 2, 3)

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Solution

The correct options are
A $(- 2, 1)$
B $(- 1, 0)$
D $(- 2, 3)$
Let $z = α$ be a real root. Then,
$α^{3} + (3 + 2 i) α + (- 1 + i a) = 0$
$\Rightarrow (α^{3} + 3 α - 1) + i (a + 2 α) = 0$
$\Rightarrow α^{3} + 3 α - 1 = 0$ and $α = - \frac{a}{2}$
$\Rightarrow - \frac{a^{3}}{8} - \frac{3 a}{2} - 1 = 0$
$\Rightarrow a^{3} + 12 a + 8 = 0$
Let $f (a) = a^{3} + 12 a + 8$ \
$∴ f (- 1) < 0, f (0) > 0, f (- 2) < 0, f (1) > 0$
and $f (3) > 0$ .
Hence, $a \in (- 1, 0)$ or $a \in (- 2, 1)$ or $a \in (- 2, 3)$

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