If the roots of the equation x^2 + x + 1 = 0 | vedclass

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

Question

(C) Given that $\alpha$ and $\beta$ are the roots of $x^2 + x + 1 = 0$,we have $\alpha + \beta = -1$ and $\alpha \beta = 1$.The roots of the equation

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(C) Given that $\alpha$ and $\beta$ are the roots of $x^2 + x + 1 = 0$,we have $\alpha + \beta = -1$ and $\alpha \beta = 1$.The roots of the equation $x^2 + px + q = 0$ are $\frac{\alpha}{\beta}$ and $\frac{\beta}{\alpha}$.By the relation between roots and coefficients,the sum of the roots is $\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = -p$.Thus,$-p = \frac{\alpha^2 + \beta^2}{\alpha \beta}$.Using the identity $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta$,we get:$-p = \frac{(\alpha + \beta)^2 - 2\alpha \beta}{\alpha \beta} = \frac{(-1)^2 - 2(1)}{1} = \frac{1 - 2}{1} = -1$.Therefore,$-p = -1$,which implies $p = 1$.

If the roots of the equation $x^2 + x + 1 = 0$ are $\alpha$ and $\beta$,and the roots of the equation $x^2 + px + q = 0$ are $\frac{\alpha}{\beta}$ and $\frac{\beta}{\alpha}$,then $p$ is equal to:

Similar Questions

The sum of the roots of a quadratic equation is $2$ and the sum of their cubes is $98$. Then the equation is:

If one root of the cubic equation $x^3-7x^2+36=0$ is double of another,then the number of negative roots is

If $\alpha+\beta=-2$ and $\alpha^3+\beta^3=-56$,then the quadratic equation whose roots are $\alpha$ and $\beta$ is

If the sum of any two roots of the equation $x^3+p x^2+q x+r=0$ is zero,then

If the roots of the quadratic equation $\frac{x - m}{mx + 1} = \frac{x + n}{nx + 1}$ are reciprocal to each other,then

Vedclass Test Series

Exam Paper Generator

Online Exam Module