Question

If a-1/a = ✓8 , then the value of a²+1/a² is

Answer 1

ATQ;

$\Longrightarrow \sf a - \dfrac{1}{a} = \sqrt{8}$

Squaring on both sides we get:

$\Longrightarrow \sf \Bigg( a - \dfrac{1}{a} \Bigg)^2 = \Big(\sqrt{8}\Big)^2$

$\Longrightarrow \sf \Bigg( a - \dfrac{1}{a} \Bigg)^2 = \sqrt{8} \times \sqrt{8}$

Use the identity (x - y)² = x² + y² - 2xy

$\Longrightarrow \sf a^2 + \Bigg(\dfrac{1}{a}\Bigg)^2 - \ 2\big(a\big)\Bigg( \dfrac{1}{a} \Bigg) = 8$

$\Longrightarrow \sf a^2 + \dfrac{1}{a^2} \ - \ 2 = 8$

$\Longrightarrow \sf a^2 + \dfrac{1}{a^2} = 8 + 2$

$\Longrightarrow \sf a^2 + \dfrac{1}{a^2} = 10$

⇒ Therefore the answer will be 10.

Answer 2

★ᏀᏆᐯᗴᑎ:-

$\sf a - \dfrac{1}{a} = \sqrt{8}$

★Ꭲᝪ ᖴᏆᑎᗞ :-

$\sf a^2 + \dfrac{1}{a^2}$

★ՏϴᏞႮͲᏆϴΝ:-

We know that,

(a-b)² = a² + b² - 2ab

Using this identity,

$\sf \Bigg ( a - \dfrac{1}{a} \Bigg )^2= (a)^2 + \Bigg ( \dfrac{1}{a} \Bigg )^2 - 2 \times \cancel{a} \times \dfrac{1}{\cancel{a}}$

$\sf\implies ( \sqrt{8} )^2= a^2 + \dfrac{1}{a^2} - 2$

$\sf \implies 8 = a^2 + \dfrac{1}{a^2} - 2$

$\sf \implies 8 +2 = a^2 + \dfrac{1}{a^2}$

$\sf \implies 10= a^2 + \dfrac{1}{a^2}$

$\boxed{ \bold{a^2 + \dfrac{1}{a^2}=10}}$

↓Answer↓

10