Let a, b in mathbb{R} and the roots alpha, be | vedclass

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

Question

(B) Given that $\alpha$ and $\beta$ are roots of the equation $z^2+az+b=0$.From the properties of roots,we have $\alpha+\beta = -a$ and $\alpha\beta =

Vedclass · Accepted Answer

$a^2=3b$

Vedclass · Answer

(B) Given that $\alpha$ and $\beta$ are roots of the equation $z^2+az+b=0$.From the properties of roots,we have $\alpha+\beta = -a$ and $\alpha\beta = b$.Since the origin $(0)$,$\alpha$,and $\beta$ form an equilateral triangle in the Argand plane,the condition for an equilateral triangle with one vertex at the origin is $\alpha^2 + \beta^2 = \alpha\beta$.We can rewrite this as $(\alpha+\beta)^2 - 2\alpha\beta = \alpha\beta$.Substituting the values of the sum and product of roots: $(-a)^2 - 2(b) = b$.This simplifies to $a^2 - 2b = b$,which gives $a^2 = 3b$.

Let $a, b \in \mathbb{R}$ and the roots $\alpha, \beta$ of the equation $z^2+az+b=0$ be complex. If the origin,$\alpha$ and $\beta$ represent the vertices of an equilateral triangle on the Argand plane,then

Similar Questions

If $|z + 4| \le 3$,then the maximum value of $|z + 1|$ is

The area of the triangle whose vertices are $i, \omega$ and $\omega^2$ is (Where $\omega$ is a complex cube root of unity other than $1$,$i$ is an imaginary number) . . . . . . sq.units

If $|z - 3 - 4i| = 4$,where $i = \sqrt{-1}$,then the maximum possible value of $|z|$ is:

The locus of the points $z$ which satisfy the condition $\text{arg} \left( \frac{z - 1}{z + 1} \right) = \frac{\pi}{3}$ is

The largest value of $r$ for which the region represented by the set $\{ \omega \in \mathbb{C} : |\omega - 4 - i| \le r \}$ is contained in the region represented by the set $\{ z \in \mathbb{C} : |z - 1| \le |z + i| \}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module