Question

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

• A. $\frac{3 -5 i}{2}$
• B. $\frac{ 5 - 3i}{2}$
• C. $\frac{3 - i}{2}$
• D. none

Answer 1

Answer: (A)

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 $\frac{-1-i}{2}$