Question

4. If x=2+√3, then
find x+1/x​

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{x+\frac{1}{x}=4}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies x = 2 + \sqrt{3} \\ \\ \red{\underline \bold{To \: Find:}} \\ \tt: \implies x + \frac{1}{x} = ?$

• According to given question :

$\bold{As \: we \: know \: that} \\ \tt: \implies {(x + \frac{1}{x}) }^{2} = {x}^{2} + \frac{1}{ {x}^{2} } + 2 \times \cancel{x} \times \frac{1}{ \cancel{x}} \\ \\ \tt: \implies {(x + \frac{1}{x}) }^{2} = {(2 + \sqrt{3}) }^{2} + \frac{1}{(2 + \sqrt{3})^{2} } + 2 \\ \\ \tt: \implies {(x + \frac{1}{x}) }^{2} = 4 + 3 + 4 \sqrt{3} + \frac{1}{4 + 3 + 4 \sqrt{3} } + 2 \\ \\ \tt: \implies {(x + \frac{1}{x}) }^{2} =7 + 4 \sqrt{3} + \frac{1}{7 + 4 \sqrt{3} } \times \frac{7 - 4 \sqrt{3} }{7 - 4 \sqrt{3} } + 2 \\ \\ \tt: \implies {(x + \frac{1}{x}) }^{2} =7 + 4 \sqrt{3} + \frac{7 - 4 \sqrt{3} }{49 - 48} + 2 \\ \\ \tt: \implies {(x + \frac{1}{x}) }^{2} =7 \cancel{+ 4 \sqrt{3}} + 7 \cancel{- 4 \sqrt{3}} + 2 \\ \\ \tt: \implies {(x + \frac{1}{x}) }^{2} = 14 + 2 \\ \\ \tt: \implies {(x + \frac{1}{x}) }^{2} = 16 \\ \\ \tt: \implies {(x + \frac{1}{x}) } = \sqrt{16} \\ \\ \green{\tt: \implies x + \frac{1}{x} =4}$

Answer 2

Given:-

$\bold{x=2+\sqrt{3}}$

To find:-

The value of $\bold{x+\frac{1}{x}}$.

Solution:-

$\bold{(x+\frac{1}{x})^2=x^2+(\frac{1}{x} )^2+2}\\\\\bold{\implies (x+\frac{1}{x})^2 =(2+\sqrt{3})^2+[\frac{1}{(2+\sqrt{3})^2 }] +2}\\\\\bold{\implies (x+\frac{1}{x})^2=4+3+4\sqrt{3}+\frac{1}{4+3+4\sqrt{3} } +2 }\\\\\bold{\implies (x+\frac{1}{x})^2=7+4\sqrt{3}+\frac{1}{7+4\sqrt{3}}+2}\\\\\bold{\implies (x+\frac{1}{x})^2=9+4\sqrt{3}+\frac{7-4\sqrt{3} }{(7-4\sqrt{3})(7+4\sqrt{3})} }\\\\\bold{\implies (x+\frac{1}{x})^2=9+4\sqrt{3}+\frac{7-4\sqrt{3} }{(7)^2-(4\sqrt{3})^2} }$

$\bold{\implies(x+\frac{1}{x})^2=9+4\sqrt3+\frac{7-4\sqrt{3} }{49-48} }\\\\\bold{\implies (x+\frac{1}{x})^2=9+4\sqrt{3}+7-4\sqrt{3} }\\\\\bold{\implies (x+\frac{1}{x})^2=9+7 }\\\\\bold{\implies (x+\frac{1}{x})^2=16}\\\\\boxed{\implies x+\frac{1}{x}=4}$

Formula(s) used:-

• (a + b)² = a² + b² + 2ab

• And rationalisation of denominator.