If alpha is a root 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

(D) Given the equation $x^2-x+1=0$. Since $\alpha$ is a root,$\alpha^2-\alpha+1=0$. Dividing by $\alpha$,we get $\alpha-1+\frac{1}{\alpha}=0$,so $\alp

Vedclass · Accepted Answer

$-9$

Vedclass · Answer

(D) Given the equation $x^2-x+1=0$. Since $\alpha$ is a root,$\alpha^2-\alpha+1=0$. Dividing by $\alpha$,we get $\alpha-1+\frac{1}{\alpha}=0$,so $\alpha+\frac{1}{\alpha}=1$. Now,$\left(\alpha+\frac{1}{\alpha}\right)^2 = \alpha^2+\frac{1}{\alpha^2}+2 = 1^2 = 1$,which implies $\alpha^2+\frac{1}{\alpha^2} = -1$. Also,$\alpha^3+\frac{1}{\alpha^3} = (\alpha+\frac{1}{\alpha})(\alpha^2+\frac{1}{\alpha^2}-1) = (1)(-1-1) = -2$. For $\alpha^4+\frac{1}{\alpha^4}$,we use $(\alpha^2+\frac{1}{\alpha^2})^2 = \alpha^4+\frac{1}{\alpha^4}+2 = (-1)^2 = 1$,so $\alpha^4+\frac{1}{\alpha^4} = -1$. Substituting these values into the expression: $(1)^3 + (-1)^3 + (-2)^3 + (-1)^3 = 1 - 1 - 8 - 1 = -9$.

If $\alpha$ is a root of the equation $x^2-x+1=0$,then $\left(\alpha+\frac{1}{\alpha}\right)^3+\left(\alpha^2+\frac{1}{\alpha^2}\right)^3+\left(\alpha^3+\frac{1}{\alpha^3}\right)^3+\left(\alpha^4+\frac{1}{\alpha^4}\right)^3=$

Similar Questions

If $\cos^4 \theta + \alpha$ and $\sin^4 \theta + \alpha$ are the roots of the equation $x^2 + 2bx + b = 0$,and $\cos^2 \theta + \beta$ and $\sin^2 \theta + \beta$ are the roots of the equation $x^2 + 4x + 2 = 0$,then find the value of $b$.

If $a, b, c$ are distinct positive real numbers and $a^2+b^2+c^2=1$,then the value of $ab+bc+ca$ is

If $\alpha, \beta, \gamma, \delta$ are the roots of the equation $x^4+x^2+1=0$,then $\frac{\alpha^3+\beta^3+\gamma^3+\delta^3}{\alpha^6+\beta^6+\gamma^6+\delta^6}=$

Let $\lambda \in R$ and let the equation $E$ be $|x|^2 - 2|x| + |\lambda - 3| = 0$. Then the largest element in the set $S = \{x + \lambda : x \text{ is an integer solution of } E\}$ is $..........$

The number of ordered pairs $(a, b)$ of positive integers such that $\frac{2a-1}{b}$ and $\frac{2b-1}{a}$ are both integers is

Vedclass Test Series

Exam Paper Generator

Online Exam Module