Question

If 1÷x=2+√3 , find the value of x2+1÷x2

Answer 1

$Given \: \frac{1}{x} = 2+\sqrt{3} \:---(1)$

$\implies x = \frac{1}{2+\sqrt{3}}\\= \frac{(2-\sqrt{3}}{(2+\sqrt{3})(2-\sqrt{3}}\\= \frac{(2-\sqrt{3}}{2^{2}-(\sqrt{3})^{2}}\\= \frac{(2-\sqrt{3}}{(4 - 3)}\\= \frac{(2-\sqrt{3}}{1}\\= 2-\sqrt{3}\:---(2)$

$x + \frac{1}{x} = 2-\sqrt{3} + 2+\sqrt{3} = 4\:--(3)$

$x^{2} + \frac{1}{x^{2}} \\= \left( x + \frac{1}{x}\right)^{2} - 2\times x \times \frac{1}{x}$

$= 4^{2} - 2 \\= 16 - 2 \\= 14$

Therefore.,

$\red {x^{2} + \frac{1}{x^{2}}}\green { = 14 }$

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