Question

if a = 2+√3 find the value of a-1/a and a²+1/a²
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Answer 1

Solution :-

a = 2 + √3

Finding the value of 1/a

1/a = 1/(2 + √3)

Rationalise the denomiator

= 1/(2 + √3) * (2 - √3)/(2 - √3)

= {2 - √3}/{2² - (√3)²}

= (2 - √3)/(4 - 3)

= (2 - √3)/1

i.e 1/a = 2 - √3

Finding the value of a - 1/a

a - 1/a = 2 + √3 - (2 - √3)

= 2 + √3 - 2 + √3

= 2√3

i.e a - 1/a = 2√3

Finding the value of a² + 1/a²

a - 1/a = 2√3

Squaring on both sides

⇒ (a - 1/a)² = (2√3)²

⇒ a² + (1/a)² - 2(a)(1/a) = 4(3)

Since (x - y)² = x² + y² - 2xy

⇒ a² + 1/a² - 2 = 12

⇒ a² + 1/a² = 12 + 2

⇒ a² + 1/a² = 14

Answer 2

Answer:-

$a {}^{2} + \frac{1}{a} = 14$

Further Explanation

$a = 2 + \sqrt{3}$

Here, we need to find value of 1/a

$= > \frac{1}{a} = \frac{1}{(2 + \sqrt{3}) }$

$= > \frac{1}{(2 + \sqrt{3} )}...... \frac{(2 - \sqrt{3}) }{(2 - \sqrt{3}) }$

$= > \frac{(2 - \sqrt{3}) }{(2 {}^{2} - \sqrt{3} {}^{2} )}$

$= > \frac{(2 - \sqrt{3}) }{(4 - 3)}$

$= > \frac{1}{a} = 2 - \sqrt{3}$

Now,

$= > \frac{a - 1}{a} = 2 + \sqrt{3} - (2 - \sqrt{3} )$

$= > 2 + \sqrt{3} - 2 + \sqrt{3}$

$= > 2 \sqrt{3} \\ = > \frac{ - 1}{a} = 2 \sqrt{3}$

Squaring both sides

$= > \frac{a - 1}{a {}^{2} } = 2 \sqrt{ 3 {}^{2} }$

$= > a {}^{2} + ( \frac{1}{a} ) {}^{2} \\ = > 2(a) \frac{1}{a} = \frac{4}{3}$

$= > a {}^{2} + \frac{1}{a} {}^{2} - 2 = 10$

$= > a {}^{2} + \frac{1}{a {}^{2} } = 12 + 2$

$= > a {}^{2} + \frac{1}{a {}^{2} } = 14$