Question 24

# If $$a-\frac{1}{a}=1$$, then $$a^2+\frac{1}{a^2}=?$$

Solution

Given

$$a-\frac{1}{a}=1$$

the

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

$$a-\frac{1}{a}=1$$ squaring both side we get

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

so we get $$a^2+\frac{1}{a^2}=3$$