Question

if x=root3+root2/root3-root2 find the value of x+1/x​

Answer 1

Solution:-

Given:-

$\sf \implies \: x = \dfrac{ \sqrt{ 3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} }$

To Find :-

$\sf \implies \: value \: of \: x + \dfrac{1}{x}$

Now Put the value

$\sf \implies \: \dfrac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } + \dfrac{1}{ \dfrac{ \sqrt{3 } + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } } \implies \: \dfrac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } + \dfrac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} }$

We get

$\sf \implies\dfrac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } + \dfrac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} }$

Now Take Lcm

$\sf\implies \: \dfrac{( \sqrt{3} + \sqrt{2})( \sqrt{3} + \sqrt{2}) + ( \sqrt{3} - \sqrt{2} )( \sqrt{3} - \sqrt{2})}{( \sqrt{3} + \sqrt{2} )( \sqrt{3} - \sqrt{2}) }$

$\sf \implies \: \dfrac{( \sqrt{3} + \sqrt{2}) ^{2} + ( \sqrt{3} - \sqrt{2}) ^{2} }{( \sqrt{3} )^{2} - ( \sqrt{2})^{2} }$

$\sf \implies \: \dfrac{( \sqrt{3} ) {}^{2} + ( \sqrt{2})^{2} + 2 \times \sqrt{3} \times \sqrt{2} +(\sqrt{3} ) {}^{2} + ( \sqrt{2})^{2} - 2 \times \sqrt{3} \times \sqrt{2} }{3 - 2}$

We get

$\sf\implies \: \dfrac{3 + 2 + 3 + 2}{1}$

$\sf \implies \: 10$

Answer is 10

Answer 2

Given:

$➪$ $\tt{x =\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}}$

To find:

$➪$ $\sf{Value\:of\:}$$\tt{x + \dfrac{1}{x}}$

Solution:

$⇒$ $\tt{x+\dfrac{1}{x}}$

$=$ $\tt{\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}+\dfrac{1}{\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}}}$

$=$ $\tt{\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}} + \dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}}$

$=$ $\tt{\dfrac{(\sqrt{3}+\sqrt{2})²+(\sqrt{3}-\sqrt{2})²}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}}$

$=$ $\tt{\dfrac{3+2+\cancel{2\sqrt{3} \sqrt{2}}+3+2-\cancel{2\sqrt{3} \sqrt{2}}}{(\sqrt{3})²-(\sqrt{2})²}}$

$=$ $\tt{\dfrac{5+5}{3-2}}$ $=$ $\tt{\dfrac{10}{1}}$$=$ $\tt{\red{10}}$

Extra information:

$\sf{•\:(a+b)²=a²+b²+2ab}$

$\sf{•\:(a-b)²=a²+b²-2ab}$

$\sf{•\:(a+b)(a-b)=a²-b²}$