Question

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

• A. $14$
• B. $21$
• C. $7$
• D. $15$

Answer 1

Answer: (C)

$7$

$a^2-3a+1=0$

$=>a^2+1=3a$

$=>a+\frac{1}{a}=3$

Now,

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

$=(3{)}^2-2$

$=7$

Thus,

$a^2+\frac{1}{a^2}=7$