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

\frac{3 - 5 i}{2}

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\frac{5 - 3 i}{2}

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\frac{3 - i}{2}

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None

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Solution

The correct option is A $\frac{3 - 5 i}{2}$
Given equation is $2 (1 + i) x^{2} - 4 (2 - i) x - 5 - 3 i = 0$
Roots $= \frac{4 (2 - i) \pm \sqrt{16 {(2 - i)}^{2} + 8 (1 + i) (5 + 3 i)}}{4 (1 + i)}$
$= \frac{4 - i}{1 + i}$ or $\frac{- i}{1 + i}$
$= \frac{3 - 5 i}{2}$ or $\frac{- 1 - i}{2}$

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Similar questions

2 (1 + i) x^{2} - 4 (2 - i) x - 5 - 3 i = 0.

i x^{2} - 2 (1 + i) x + (2 - i) = 0

(3 - i)

i = \sqrt{- 1}

(2 + i) x^{2} - (5 - i) x + 2 (1 - i) = 0

\frac{(1 + 3 i) (2 - 5 i)}{(2 - \sqrt{6} i) (- 3 + 2 \sqrt{5} i)}

Solve the following quadratic equations :

(i) $x^{2} - (3 \sqrt{2} + 2 i) x + 6 \sqrt{2} i = 0$

(ii) $x^{2} - (5 - i) x + (18 + i) = 0$

(iii) $(2 + i) x^{2} - (5 - i) x + 2 (1 - i) = 0$

(iv) $x^{2} - (2 + i) x - (1 - 7 i) = 0$

(v) $i x^{2} - 4 x - 4 i = 0$

(vi) $x^{2} + 4 i x - 4 = 0$

(vii) $2 x^{2} + \sqrt{15} i x - i = 0$

(viii) $x^{2} - x + (1 + i) = 0$

(ix) $i x^{2} - x + 12 i = 0$

(x) $x^{2} - (3 \sqrt{2} - 2 i) x - \sqrt{2} i = 0$

(xi) $x^{2} - (\sqrt{2} + i) x + \sqrt{2} i = 0$

(xii) $2 x^{2} - (3 + 7 i) x + (9 i - 3) = 0$

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