If $a^2-6a-1=0$, find the value of $a^2 +\dfrac {1}{a^2}$

Question

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?

Ok, I found it. It is 38. – uniquesolution Apr 15 '17 at 13:28 — uniquesolution, Apr 15 '17 at 13:28
Divide both sides of the condition by $a$ – lab bhattacharjee Apr 15 '17 at 13:28 — lab bhattacharjee, Apr 15 '17 at 13:28
@uniquesolution, could you show the calculations? – pi-π Apr 15 '17 at 13:29 — pi-π, Apr 15 '17 at 13:29
Too late. They beat me to it. – uniquesolution Apr 15 '17 at 13:29 — uniquesolution, Apr 15 '17 at 13:29

score 6 · Accepted Answer · answered Apr 15 '17 at 13:28

6

Your equation gives you (move the linear term to the other side and divide by $a$) $$a-\frac1a=6.$$ Squaring this gives. Oops, you do it.

answered Apr 15 '17 at 13:28

Jyrki Lahtonen

1

Community Wikifying this, because I have already used the same technique earlier. – Jyrki Lahtonen Apr 15 '17 at 13:31

score 1 · Answer 2 · answered Apr 15 '17 at 13:28

1

You have $$a-\frac 1a=6$$ so squaring gives $$a^2-2+\frac{1}{a^2}=36$$ so...?

answered Apr 15 '17 at 13:28

David Quinn

score 0 · Answer 3 · edited May 31 '22 at 23:49

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.

3 Answers3