If x+frac{1}{x}=3, then the value of x^{8}+fr | vedclass

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

Question

(D) Given $x+\frac{1}{x}=3.$Squaring both sides,we get $\left(x+\frac{1}{x}\right)^{2}=3^{2} \Rightarrow x^{2}+\frac{1}{x^{2}}+2=9 \Rightarrow x^{2}+\

Vedclass · Accepted Answer

$None \, of \, these$

Vedclass · Answer

(D) Given $x+\frac{1}{x}=3.$Squaring both sides,we get $\left(x+\frac{1}{x}\right)^{2}=3^{2} \Rightarrow x^{2}+\frac{1}{x^{2}}+2=9 \Rightarrow x^{2}+\frac{1}{x^{2}}=7.$Now,squaring again,$\left(x^{2}+\frac{1}{x^{2}}\right)^{2}=7^{2} \Rightarrow x^{4}+\frac{1}{x^{4}}+2=49 \Rightarrow x^{4}+\frac{1}{x^{4}}=47.$Finally,squaring once more,$\left(x^{4}+\frac{1}{x^{4}}\right)^{2}=47^{2} \Rightarrow x^{8}+\frac{1}{x^{8}}+2=2209 \Rightarrow x^{8}+\frac{1}{x^{8}}=2207.$

If $x+\frac{1}{x}=3,$ then the value of $x^{8}+\frac{1}{x^{8}}$ will be

Similar Questions

If $\alpha, \beta, \gamma$ are roots of the equation $x^3 + ax^2 + bx + c = 0$,then $\alpha^{-1} + \beta^{-1} + \gamma^{-1} = $

If the roots of the equation ${x^2} - 8x + ({a^2} - 6a) = 0$ are real,then

The number of integral values of $a$ for which $x^2 - (a - 1)x + 3 = 0$ has both roots positive and $x^2 + 3x + 6 - a = 0$ has both roots negative is

If one root of the quadratic equation $2x^{2} + Px + 4 = 0$ is $2$,find the second root and the value of $P$.

If $\alpha$ and $\beta$ are the roots of the equation $ax^2 + bx + c = 0$,then the equation whose roots are $\alpha + \frac{1}{\beta}$ and $\beta + \frac{1}{\alpha}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module