Question

If $a-\frac{1}{a}=3$, find $a^2+3a+\frac{1}{a^2}-\frac{3}{a}$.

• A. $20$
• B. $12$
• C. $32$
• D. $17$

Answer 1

The correct option is A $20$

Given,
$a-\frac{1}{a}=3$
Squaring both sides,
$=>(a-\frac{1}{a}{)}^2=3^2$
$=>a^2+\frac{1}{a^2}-2\left(a\right)\left(\frac{1}{a}\right)=9$
$=>a^2+\frac{1}{a^2}=9+2$
$=>a^2+\frac{1}{a^2}=11$
now,
$a^2+3a+\frac{1}{a^2}-\frac{3}{a}$
$=a^2+\frac{1}{a^2}+3a+-\frac{3}{a}$
$=a^2+\frac{1}{a^2}+3(a-\frac{1}{a})$
$=11+3\left(3\right)$
$=11+9$
$=20$