First slide

Theory of equations

Question

The root of the equation $2 (1 + i) x^{2} - 4 (2 - i) x - 5 - 3 i = 0, where i = \sqrt{- 1},$ which has greater modulus, is

Moderate

A

$\frac{3 - 5 i}{2}$
B

$\frac{5 - 3 i}{2}$
C

$\frac{3 + i}{2}$
D

$\frac{3 i + 1}{2}$

Solution

The given equation is
$2 (1 + i) x^{2} - 4 (2 - i) x - 5 - 3 i = 0$
$\Rightarrow x = \frac{4 (2 - i) \pm \sqrt{16 (2 - i)^{2} + 8 (1 + i) (5 + 3 i)}}{4 (1 + i)} = \frac{4 (2 - i) \pm \sqrt{48 - 64 i + 40 - 24 + 64 i}}{4 (1 + i)} = \frac{4 (2 - i) \pm \sqrt{64}}{4 (1 + i)} = \frac{8 - 4 i \pm 8}{4 (1 + i)} = - \frac{i}{1 + i} or \frac{4 - i}{1 + i} = \frac{- 1 - i}{2} or \frac{3 - 5 i}{2}$
Now, $|\frac{- 1 - i}{2}| = \sqrt{\frac{1}{4} + \frac{1}{4}} = \sqrt{\frac{1}{2}}$
$and |\frac{3 - 5 i}{2}| = \sqrt{\frac{9}{4} + \frac{25}{4}} = \sqrt{\frac{17}{2}}$
Also, $\sqrt{\frac{17}{2}} > \sqrt{\frac{1}{2}}$
Hence, required root is $\frac{3 - 5 i}{2}$

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