Question

If $a^2-3a+1=0$ find $a^2+\frac{1}{a^2}$

• A. $8$
• B. $13$
• C. $7$
• D. $2$

Answer 1

Answer: (C)

$7$

$a^2-3a+1=0$
$=>a^2+1=3a$
$=>a+\frac{1}{a}=3$
Squaring both the sides,
$(a+\frac{1}{a}{)}^2=3^2$
$a^2+\left(\frac{1}{a^2}\right)+2\times a\times \frac{1}{a}=9$
$a^2+\frac{1}{a^2}=9-2$
$a^2+\frac{1}{a^2}=7$