Let alpha and beta be the roots of the equati | vedclass

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

Question

(D) The given equation is $x^2 + x + 1 = 0$.The roots of this equation are the complex cube roots of unity,$\omega$ and $\omega^2$.Let $\alpha = \omeg

Vedclass · Accepted Answer

$x^2 + x + 1 = 0$

Vedclass · Answer

(D) The given equation is $x^2 + x + 1 = 0$.The roots of this equation are the complex cube roots of unity,$\omega$ and $\omega^2$.Let $\alpha = \omega$ and $\beta = \omega^2$.We need to find the equation whose roots are $\alpha^{19}$ and $\beta^7$.Since $\omega^3 = 1$,we have $\alpha^{19} = \omega^{19} = (\omega^3)^6 \cdot \omega = 1^6 \cdot \omega = \omega$.Similarly,$\beta^7 = (\omega^2)^7 = \omega^{14} = (\omega^3)^4 \cdot \omega^2 = 1^4 \cdot \omega^2 = \omega^2$.The roots of the new equation are $\omega$ and $\omega^2$.Therefore,the required equation is the same as the original equation: $x^2 + x + 1 = 0$.

Let $\alpha$ and $\beta$ be the roots of the equation $x^2 + x + 1 = 0$. The equation whose roots are $\alpha^{19}$ and $\beta^7$ is:

Similar Questions

If $a+b+c=0,$ then find the value of $\frac{1}{b^{2}+c^{2}-a^{2}}+\frac{1}{c^{2}+a^{2}-b^{2}}+\frac{1}{a^{2}+b^{2}-c^{2}}.$

The equation $e^x - x - 1 = 0$ has

Two positive distinct numbers $a$ and $b$ each differ from their reciprocal by $1$. The value of $a + b$ is

If $\left\{\frac{1}{2}(a-b)\right\}^{2}+a b=p(a+b)^{2},$ then the value of $p$ is (assume that $a \neq-b).$

Let $[t]$ denote the greatest integer $\leq t$. Then the equation in $x$,$[x]^{2} + 2[x + 2] - 7 = 0$,has

Vedclass Test Series

Exam Paper Generator

Online Exam Module