Let {z_1} and {z_2} be two roots of the equat | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given that ${z_1}$ and ${z_2}$ are the roots of the equation ${z^2 + az + b = 0}$.By Vieta's formulas,we have ${z_1 + z_2 = -a}$ and ${z_1 z_2 = b

Vedclass · Accepted Answer

${a^2 = 3b}$

Vedclass · Answer

(C) Given that ${z_1}$ and ${z_2}$ are the roots of the equation ${z^2 + az + b = 0}$.By Vieta's formulas,we have ${z_1 + z_2 = -a}$ and ${z_1 z_2 = b}$.Since the origin $({z_3 = 0})$,${z_1}$,and ${z_2}$ form an equilateral triangle,the condition for the vertices ${z_1, z_2, z_3}$ to form an equilateral triangle is ${z_1^2 + z_2^2 + z_3^2 = z_1 z_2 + z_2 z_3 + z_3 z_1}$.Substituting ${z_3 = 0}$,we get ${z_1^2 + z_2^2 = z_1 z_2}$.Adding ${2 z_1 z_2}$ to both sides,we get ${z_1^2 + z_2^2 + 2 z_1 z_2 = 3 z_1 z_2}$.This simplifies to ${(z_1 + z_2)^2 = 3 z_1 z_2}$.Substituting the values from Vieta's formulas,we get ${(-a)^2 = 3b}$,which implies ${a^2 = 3b}$.

Let ${z_1}$ and ${z_2}$ be two roots of the equation ${z^2 + az + b = 0}$,where ${z}$ is a complex number. Further,assume that the origin,${z_1}$,and ${z_2}$ form an equilateral triangle. Then:

Similar Questions

The number of values of $z \in \mathbb{C}$, satisfying the equations $|z - (4 + 8i)| = \sqrt{10}$ and $|z - (3 + 5i)| + |z - (5 + 11i)| = 4\sqrt{5}$, is:

The complex numbers $z_1, z_2, z_3$ are the vertices of a triangle. Then the complex numbers $z$ which make the triangle into a parallelogram are

The locus of $z$ given by $\left| \frac{z - 1}{z - i} \right| = 1$ is

Let $S = \{z = x + iy : |z - 1 + i| \geq |z|, |z| < 2, |z + i| = |z - 1|\}$. Then the set of all values of $x$,for which $w = 2x + iy \in S$ for some $y \in \mathbb{R}$,is:

The inequality $|z - 4| < |z - 2|$ represents the region given by

Vedclass Test Series

Exam Paper Generator

Online Exam Module