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 the figure shows the graph of $y = ax^2 + bx + c$,then . . . . . .

If the lengths of two sides of a triangle are the roots of the equation $x^2-2 \sqrt{3} x+2=0$ and the angle between these sides is $\frac{\pi}{3}$,then the perimeter of the triangle is

If $x$ is real,the maximum value of $\frac{3x^2 + 9x + 17}{3x^2 + 9x + 7}$ is

Let $a, b, c, d$ be distinct real numbers such that $a, b$ are roots of $x^2-5cx-6d=0$,and $c, d$ are roots of $x^2-5ax-6b=0$. Then $b+d$ is

Let $p$ and $q$ be two real numbers such that $p+q=3$ and $p^{4}+q^{4}=369$. Then $\left(\frac{1}{p}+\frac{1}{q}\right)^{-2}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module