Question

If $x=3+\sqrt{8}$

then $x^4+\frac{1}{x^4}=$ ______

• A. 1154
• B. 1156
• C. 36
• D. 34

Answer 1

The correct option is A

1154

$x=3+\sqrt{8}$ ...(i)

$\frac{1}{x}=\frac{1}{3+\sqrt{8}}\times \frac{3-\sqrt{8}}{3-\sqrt{8}}$

=$\frac{3-\sqrt{8}}{(3{)}^2-(\sqrt{8}{)}^2}$ =$\frac{3-\sqrt{8}}{9-8}$

=$3-\sqrt{8}$ ...(ii)

$\therefore x+\frac{1}{x}=3+\sqrt{8}+3-\sqrt{8}$ (from (i) and (ii))

= $6$

$(x+\frac{1}{x}{)}^2=x^2+2+\frac{1}{x^2}$

$(6{)}^2=x^2+2+\frac{1}{x^2}$

$x^2+\frac{1}{x^2}=34$

$(x^2+\frac{1}{x^2}{)}^2=x^4+2+\frac{1}{x^4}$

$\therefore x^4+\frac{1}{x^4}=(34{)}^2-2$

=$1156-2$

=$1154$