Question

if x=√3-√2 , find the value of:
(i) x+1/x
​

Answer 1

AnSwEr

${\tt{\red{\underline{\large{Given}}}}}$

• X=√3-√2

${\sf{\green{\underline{\large{To\:find}}}}}$

• X + 1/X

${\sf{\pink{\underline{\Large{Explanation}}}}}$

• Finding the value of 1/X
• we have to rationalize the given term

Let do this

$\tt\implies\frac{1}{X}=\dfrac{1}{\sqrt{3}-\sqrt{2}}$

$\tt\implies\frac{1}{X}=\dfrac{1}{\sqrt{3}-\sqrt{2}}\times\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}$

=>1/X=√3+√2 / 1

now adding the value

=>X+1/X = √3-√2+√3+√2

=>X+1/X = (√3-√2)+√3+√2

=>X+1/X =√3-√2+√3+√2 }

=>X+1/X =2√3

Answer 2

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→ Answer:

x + 1/x = 2√3

→ Step-by-step explanation:

$x=\sqrt{3}-\sqrt{2}\\\\\frac{1}{x}=\frac{1}{\sqrt{3}-\sqrt{2}}$

Rationalize the denominator.

Rationalizing factor of $\sqrt{3}-\sqrt{2}$ is $\sqrt{3}+\sqrt{2}$

$=>\frac{1}{x}=\frac{1}{\sqrt{3}-\sqrt{2}} *\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}} \\\\=>\frac{1}{x}=\frac{\sqrt{3}+\sqrt{2}}{3-2}\\\\ =>\frac{1}{x}=\sqrt{3}+\sqrt{2}$

$So,x+\frac{1}{x}=\sqrt{3}-\sqrt{2}+\sqrt{3}+\sqrt{2} \\\\=>x+\frac{1}{x}=2\sqrt{3}$

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