Question:medium

The root of the equation 2(1+i)x²-4(2-i)x-5-3i=0 which has greater modulus is

Show Hint

After finding roots, compare |z| to determine the larger modulus.

Updated On: Mar 20, 2026

\( \dfrac{3-5i}{2} \)
\( \dfrac{5-3i}{2} \)
\( \dfrac{3-i}{2} \)
none

Show Solution

The Correct Option is B

Solution and Explanation

Was this answer helpful?

0

Top Questions on Complex Numbers and Quadratic Equations

If \(|z_1| = 2, |z_2| = 3, |z_3| = 4\) and \(|2z_1 + 3z_2 + 4z_3| = 4\), then absolute value of \(8z_2z_3 + 27z_3z_1 + 64z_1z_2\) equals
- BITSAT - 2026
- Mathematics
- Complex Numbers and Quadratic Equations
View Solution
If \( |z_1| = 2 \), \( |z_2| = 3 \), \( |z_3| = 4 \) and \( |2z_1 + 3z_2 + 4z_3| = 4 \), then absolute value of \( 8z_2 z_3 + 27z_3 z_1 + 64z_1 z_2 \) equals
- BITSAT - 2026
- Mathematics
- Complex Numbers and Quadratic Equations
View Solution
If $50 \left( \frac{2x}{1 + 3i} + \frac{y}{1 - 2i} \right) = 31 + 17i$ where $x, y \in R$ & $i = \sqrt{-1}$ then value of $10(x + 3y)$ is ________
- JEE Main - 2026
- Mathematics
- Complex Numbers and Quadratic Equations
View Solution
If $^{36}C_{r+1} = 6 \cdot \frac{^{35}C_r}{k^2 - 3}$, then number of ordered pairs $(r, k)$, where $r, k \in Z$ is ______.
- JEE Main - 2026
- Mathematics
- Complex Numbers and Quadratic Equations
View Solution
Want to practice more? Try solving extra ecology questions todayView All Questions

Questions Asked in BITSAT exam

The locus of the mid-point of a chord of the circle \( x^2 + y^2 = 4 \), which subtends a right angle at the origin is

BITSAT - 2026
circle

A passenger sitting in a train A moving at 90 km/h observes another train B moving in the opposite direction for 8 s. If the velocity of the train B is 54 km/h, then length of train B is:

BITSAT - 2026
Relative Velocity

If \( x = \sqrt{2^{\text{cosec}^{-1} t}} \) and \( y = \sqrt{2^{\text{sec}^{-1} t}} (|t| \ge 1) \), then dy/dx is equal to :

BITSAT - 2026
Derivatives of Functions in Parametric Forms

If \( f(x) = \int_0^x t(\sin x - \sin t)dt \) then :

BITSAT - 2026
Definite Integral