If 2+4i is one of the roots of x^2+bx+c=0 wit | vedclass

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

Question

(D) Given that $2+4i$ is one of the roots of the quadratic equation $x^2+bx+c=0$ where $b, c \in R$. Since the coefficients are real,the complex roots

Vedclass · Accepted Answer

$(-4, 20)$

Vedclass · Answer

(D) Given that $2+4i$ is one of the roots of the quadratic equation $x^2+bx+c=0$ where $b, c \in R$. Since the coefficients are real,the complex roots must occur in conjugate pairs. Therefore,the other root is $2-4i$. Sum of roots $= -b = (2+4i) + (2-4i) = 4$. Thus,$b = -4$. Product of roots $= c = (2+4i)(2-4i) = 2^2 - (4i)^2 = 4 + 16 = 20$. Thus,$c = 20$. Therefore,$(b, c) = (-4, 20)$. Hence,option $(d)$ is correct.

If $2+4i$ is one of the roots of $x^2+bx+c=0$ with $b, c \in R$,then $(b, c) =$

Similar Questions

If the roots of $4x^2 + px + 9 = 0$ are equal,then the absolute value of $p$ is:

If one root of the equation $x^2 + px + q = 0$ is $2 + \sqrt{3}$,then the values of $p$ and $q$ are:

The number of real values of $a$ satisfying the equation $a^2 - 2a\sin x + 1 = 0$ is

The number of real solutions of the equation $|x^2 + 4x + 3| + 2x + 5 = 0$ is:

What are the roots of the equation $x^{2/3} + x^{1/3} - 2 = 0$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module