Question

If x=3 + 2root2 find 1/x and x-1/x

Answer 1

Answer:

>> Given x = 3 - 2√2

>> x - 1/x

>>  3+2√2 - [1 / 3+2√2]

>>  3+2√2 - [3-2√2 / 3+2√2(3-2√2)]

>>  3+2√2 - [3-2√2 / 3² - (2√2)²]

>>  3+2√2 - (3-2√2/9-8)

>> 3+2√2 - (3 - 2√2)

>> 4√2

>> x - 1/x = 4√2

>> 1/x = 3 - 2√2

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Answer 2

Answer

→ ¹/ₓ = 3 - 2√2

→ x - ¹/ₓ = 4√2

Given

$\bullet \; \; \rm x=3+2\sqrt{2}$

To Find

$\bullet \; \; \rm \dfrac{1}{x}\\\\\bullet \;\; \rm x-\dfrac{1}{x}$

Solution

$\rm x=3+2\sqrt{2}\ \; \blue{\bigstar}\\\\\to \rm \dfrac{1}{x}=\dfrac{1}{3+2\sqrt{2}}\\\\\rm Rationalize\ the\ denominator\\\\\to \rm \dfrac{1}{x}=\dfrac{1}{3+2\sqrt{2}}\times \dfrac{3-2\sqrt{2}}{3-2\sqrt{2}}\\\\\to \rm \dfrac{1}{x}=\dfrac{3-2\sqrt{2}}{(3)^2-(2\sqrt{2})^2}\ \\\\ \bf{\because\ (a+b)(a-b)=a^2-b^2}\\\\\to \rm \dfrac{1}{x}=\dfrac{3-2\sqrt{2}}{9-8}\\\\\to \rm \dfrac{1}{x}=3-2\sqrt{2}\ \; \pink{\bigstar}$

Now , Let's find our required value ,

$\to \rm x-\dfrac{1}{x}\\\\\to \rm \{(3+2\sqrt{2})-(3-2\sqrt{2})\}\\\\\to \rm 2\sqrt{2}+2\sqrt{2}\\\\\to \rm \orange{\bigstar}\ \; 4\sqrt{2}\ \; \green{\bigstar}$