Question

If $a^2-4a-1=0,a\ne 0$, then the value of $a^2+3a+\frac{1}{a^2}-\frac{3}{a}$ is

• A. $24$
• B. $26$
• C. $28$
• D. $30$

Answer 1

The correct option is D $30$

$a^2-4a-1=0$
$\Rightarrow \frac{a^2-1}{a}=4\Rightarrow a-\frac{1}{a}=4$
Now, $a^2+3a+\frac{1}{a^2}=\frac{3}{a}$
$=a^2+\frac{1}{a^2}+3(a-\frac{1}{a})={(a-\frac{1}{a})}^2+2+3(a-\frac{1}{a})$
$=4^2+2+3\times 4=30$.