Question

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

• A. $194$
• B. $192$
• C. $196$
• D. $190$

Answer 1

The correct option is A $194$

Given,

$x+\frac{1}{x}=4$

Taking square on both sides,

${(x+\frac{1}{x})}^2=4^3$

Using Identity,

$(a+b{)}^2=a^2+2ab+b^2$

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

${\left(\frac{1}{x}\right)}^2=16$

$x^2+\frac{1}{x^2}=16-2$

$\therefore x^2+\frac{1}{x^2}=14$

Again, taking square on both sides,

${(x^2+\frac{1}{x^2})}^2={14}^2$

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

$\frac{1}{x^2}+{\left(\frac{1}{x^2}\right)}^2=196$

$x^4+2+\frac{1}{x^4}=196$

$x^4+\frac{1}{x^4}=196-2$

$\therefore x^4+\frac{1}{x^4}=194$

Correct Option is $b$.