Question

if a-1/a=2find the value of a²+1/a²​

Answer 1

Required Answer:-

Given:

• a - 1/a = 2

To find:

• The value of a² + 1/a²

Solution:

Given that,

$\rm \implies a - \dfrac{1}{a} = 2$

Squaring both sides,, we get,

$\rm \implies { \bigg(a - \dfrac{1}{a} \bigg)}^{2} = 2$

$\rm \implies {a}^{2} + \dfrac{1}{ {a}^{2} } - 2 \times a \times \dfrac{1}{a} = {2}^{2}$

$\rm \implies {a}^{2} + \dfrac{1}{ {a}^{2} } - 2 = 4$

$\rm \implies {a}^{2} + \dfrac{1}{ {a}^{2} } = 4 + 2$

$\rm \implies {a}^{2} + \dfrac{1}{ {a}^{2} } = 6$

Hence, the value of a² + 1/a² is 6.

Answer:

$\rm \implies {a}^{2} + \dfrac{1}{ {a}^{2} } = 6$

Identity Used:

• (a + b)² = a² + 2ab + b²

Other Identities:

• (a - b)² = a² - 2ab + b²
• a² - b² = (a + b)(a - b)
• (a + b)² = (a - b)² + 4ab
• (a - b)² = (a + b)² - 4ab

Answer 2

$\sf{a-\dfrac{1}{a}=2}$

We are All Aware of Algebraic Identity : (a - b)² = a² + b² - 2ab

To Use This Identity in our Question, We need to Square on Both Sides

$\to\sf{\left(a-\dfrac{1}{a}\right)^2=2^2}$

Now, It's Time For Identify :D

$\to\sf{\left(a^2+\dfrac{1^2}{a^2}-2\times a\times \dfrac{1}{a}\right)=2^2}$

$\to\sf{\left(a^2+\dfrac{1}{a^2}-2\right)=2^2}$

$\to\sf{a^2+\dfrac{1}{a^2}-2=4}$

Adding 2 to Both Sides

$\to\sf{a^2+\dfrac{1}{a^2}-2+2=4+2}$

$\to\sf{a^2+\dfrac{1}{a^2}=6}$