<s>[INST] If $a^2-6a-1=0$, find the value of $a^2 +\dfrac {1}{a^2}$ If $a^2-6a-1=0$, find the value of $a^2 +\dfrac {1}{a^2}$
My Attempt:
$$a^2+\dfrac {1}{a^2}=(a+\dfrac {1}{a})^2 - 2a.\dfrac {1}{a}$$
$$a^2+\dfrac {1}{a^2}=(a+\dfrac {1}{a})^2 - 2$$
How do I proceed further?
 [/INST] This can be done by using Sridharacharya Formula.

The roots of a quadratic equation $ax^2+bx+c$ are $$\boxed{\frac{-b \pm \sqrt{b^2-4ac}}{2a}}$$

Thus,
$\Large a=\frac{6 \pm \sqrt{6^2+4}}{2}$
$\large a=3 \pm \sqrt{10}$
Now,  $$a^2+a^{-2}=\bigg(a+\frac1a\bigg)^2-2$$
Now, one can put both the values of $a$, and then by rationalizing the denominator we get answer as $\boxed{38}$ in both the case.
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