Question

what is the value of a×a+1/a×a​

Answer 1

Given :-

$\: \: \: \: \bullet \: \rm \: a + \dfrac{1}{a} = \dfrac{1}{ \sqrt{3} - 1 }$

To find:- Value of a² + 1/a²

‎‎‎‎‎‎‎ㅤㅤㅤㅤㅤㅤㅤㅤㅤ__________________________

Solution:-

ㅤㅤㅤㅤ❍Let's do squaring on both sides for the given equation.

$\: \: \: \mapsto \: \rm \: \bigg(a + \dfrac{1}{a} \bigg) {}^{2} = \bigg( \dfrac{1}{ \sqrt{3} - 1 } \bigg) {}^{2}$

☆ Expanding using formulae :-

• (a + b)² = a² + 2ab + b²

• (a -b )² = a² -2ab + b²

$\: \: \: \: \mapsto \rm \: a{}^{2} + \dfrac{1}{a {}^{2} } + 2.a. \dfrac{1}{a} = \dfrac{1}{ (\sqrt{3} - 1) {}^{2} }$

$\: \: \: \: \mapsto \rm \: a{}^{2} + \dfrac{1}{a {}^{2} } + 2 = \dfrac{1}{ (\sqrt{3} ) {}^{2} + (1) {}^{2} - 2( \sqrt{3} )(1) }$

$\: \: \: \: \mapsto \rm \: a{}^{2} + \dfrac{1}{a {}^{2} } + 2 = \dfrac{1}{ 4 - 2 \sqrt{3} }$

$\: \: \: \: \: \mapsto \: \rm \: a {}^{2} + \dfrac{1}{a {}^{2} } = \dfrac{1}{4 - 2 \sqrt{3} } - 2$

$\: \: \: \: \: \: \mapsto \rm \: a {}^{2} + \dfrac{1}{a {}^{2} } = \dfrac{1 - 2(4 - 2 \sqrt{3} )}{4 - 2 \sqrt{3} }$

$\: \: \: \: \: \: \mapsto \rm \: a {}^{2} + \dfrac{1}{a {}^{2} } = \dfrac{1 - 8 + 4 \sqrt{3} }{4 - 2 \sqrt{3} }$

☆ Multiplying with its Rationalizing factor of denominator.

$\: \: \: \: \: \: \mapsto \rm \: a {}^{2} + \dfrac{1}{a {}^{2} } = \dfrac{ - 7 + 4 \sqrt{3} }{4 - 2 \sqrt{3} } \times \dfrac{4 + 2 \sqrt{3} }{4 + 2 \sqrt{3} }$

$\: \: \: \: \: \: \mapsto \rm \: a {}^{2} + \dfrac{1}{a {}^{2} } = \dfrac{ - 7 (4 + 2 \sqrt{3}) + 4 \sqrt{3} (4 + 2 \sqrt{3}) }{(4) {}^{2} - 12 }$

$\: \: \: \: \: \: \mapsto \rm \: a {}^{2} + \dfrac{1}{a {}^{2} } = \dfrac{ -28 - 14 \sqrt{3} + 16 \sqrt{3} + 24 }{4 }$

$\: \: \: \: \: \: \mapsto \rm \: a {}^{2} + \dfrac{1}{a {}^{2} } = \dfrac{ -4 + 2 \sqrt{3} }{4 }$

$\: \: \: \: \: \:\mapsto \rm \: a {}^{2} + \dfrac{1}{a {}^{2} } = \cancel\dfrac{ -4}{ 4} + \cancel\dfrac{2 \sqrt{3} }{4}$

$\: \: \: \: \: \: \mapsto \: \: \: \rm \: - 1 + \dfrac{ \sqrt{3} }{2} \pink \bigstar$

So, the correct answer is Option-B