Question

If $a+\frac{1}{a}=2$, then find $a^4+\frac{1}{a^4}$.

Answer 1

Applying the formula, ${(a+b)}^2=a^2+b^2+2ab$

${(a+\frac{1}{a})}^2=a^2+{\left(\frac{1}{a}\right)}^2+2a\times \frac{1}{a}=a^2+\frac{1}{a^2}+2$

$=>a^2+\frac{1}{a^2}={(a+\frac{1}{a})}^2-2$
Substituting $a+\frac{1}{a}=2$,

$=>a^2+\frac{1}{a^2}=2^2-2=4-2=2$

Squaring both sides,

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

Applying the formula, ${(a+b)}^2=a^2+b^2+2ab$

$a^4+\frac{1}{a^4}+2\times a^2\times \frac{1}{a^2}=4$

$a^4+\frac{1}{a^4}+2=4$
$a^4+\frac{1}{a^4}=2$