If alpha and beta are the roots of the equati | vedclass

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

Question

(C) The given quadratic equation is $x^2-2x+4=0$. Since $\alpha$ and $\beta$ are the roots,we have $\alpha+\beta=2$ and $\alpha\beta=4$. Note that the

Vedclass · Accepted Answer

$-2^{10}$

Vedclass · Answer

(C) The given quadratic equation is $x^2-2x+4=0$. Since $\alpha$ and $\beta$ are the roots,we have $\alpha+\beta=2$ and $\alpha\beta=4$. Note that the roots of $x^2-2x+4=0$ are $x = \frac{2 \pm \sqrt{4-16}}{2} = 1 \pm i\sqrt{3}$. In polar form,$1 \pm i\sqrt{3} = 2(\cos(\frac{\pi}{3}) \pm i\sin(\frac{\pi}{3})) = 2e^{\pm i\pi/3}$. Thus,$\alpha = 2e^{i\pi/3}$ and $\beta = 2e^{-i\pi/3}$. Then $\alpha^9 = (2e^{i\pi/3})^9 = 2^9 e^{i3\pi} = 2^9(\cos(3\pi) + i\sin(3\pi)) = 2^9(-1) = -2^9$. Similarly,$\beta^9 = (2e^{-i\pi/3})^9 = 2^9 e^{-i3\pi} = 2^9(\cos(-3\pi) + i\sin(-3\pi)) = 2^9(-1) = -2^9$. Therefore,$\alpha^9+\beta^9 = -2^9 - 2^9 = -2 	imes 2^9 = -2^{10}$.

If $\alpha$ and $\beta$ are the roots of the equation $x^2-2x+4=0$,then $\alpha^9+\beta^9$ is equal to

Similar Questions

Rectified spirit is a mixture of

Four resistors $A, B, C$ and $D$ form a Wheatstone bridge as shown in the figure. The bridge is balanced when $C = 100 \Omega$. If $A$ and $B$ are interchanged,the bridge balances for $C = 121 \Omega$. The value of $D$ is (in $\Omega$)

If $y=a e^x+b e^{-x}+c$,where $a, b, c$ are parameters,then $y^{\prime \prime \prime}$ is equal to

In $\triangle ABC$,$\frac{r_1-r}{a}+\frac{r_2-r}{b}+\frac{r_3-r}{c} =$

It is estimated that per minute each $cm^2$ of the Earth receives about $2 \ cal$ $(1 \ cal = 4.18 \ J)$ of heat energy from the Sun. This is called the Solar constant. In $SI$ units,the value is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module