Question

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

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

Answer 1

Answer: (A)

Roots $= \frac{4\left(2-i\right)\pm\sqrt{16\left(2-i\right)^{2}+8\left(1+i\right)\left(5+3i\right)}}{4\left(1+i\right)}$
$= \frac{4-i}{1+i}$ or $\frac{-i}{1+i} = \frac{3-5i}{2}$ or $\frac{-1-i}{2}$
$= \left|\frac{3-5i}{2}\right| = \sqrt{\frac{9+25}{4}} = \sqrt{\frac{17}{2}}$ and $\left|\frac{-1-i}{2}\right| = \sqrt{\frac{1+1}{4}} = \sqrt{\frac{1}{2}}$