Question

If $\mathtt{x = 1 - \sqrt2}$, then Find the Value of $\mathtt{\Big(x -\dfrac{1}{x}\Big)^2}$​

Answer 1

Solution -

We have,

• $\sf{x = 1 - \sqrt{2}}$

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To find :-

• $\sf{\bigg( x - \dfrac{1}{x} \bigg)^2}$

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Firstly, we will solve the bracket.

$\tt:\implies\: \: \: \: \: \: \: \: {\bigg( \dfrac{x^2 - 1}{x} \bigg)^2}$

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Now, put the value of x.

$\tt:\implies\: \: \: \: \: \: \: \: {\bigg( \dfrac{(1 - \sqrt{2})^2 - 1}{1 - \sqrt{2}} \bigg)^2}$

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$\tt:\implies\: \: \: \: \: \: \: \: {\bigg( \dfrac{1 + 2 - 2 \sqrt{2} - 1}{1 - \sqrt{2}} \bigg)^2}$

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$\tt:\implies\: \: \: \: \: \: \: \: {\bigg( \dfrac{2 - 2\sqrt{2}}{1 - \sqrt{2}} \bigg)^2}$

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We will open the brackets now by squaring the whole term.

$\tt:\implies\: \: \: \: \: \: \: \: {\dfrac{(2 - 2\sqrt{2})^2}{(1 - \sqrt{2})^2}}$

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Using identity

• (a - b)² = a² + b² - 2ab

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$\tt:\implies\: \: \: \: \: \: \: \: {\dfrac{(2)^2 + (2\sqrt{2})^2 - 8\sqrt{2}}{(1)^2 + (\sqrt{2})^2 - 2\sqrt{2}}}$

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$\tt:\implies\: \: \: \: \: \: \: \: {\dfrac{4 + 8 - 8 \sqrt{2}}{1 + 2 - 2 \sqrt{2}}}$

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$\tt:\implies\: \: \: \: \: \: \: \: {\dfrac{12 - 8\sqrt{2}}{3 - 2\sqrt{2}}}$

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Rationalising the denominator

$\tt:\implies\: \: \: \: \: \: \: \: {\dfrac{12 - 8\sqrt{2}}{3 - 2\sqrt{2}} \times \dfrac{3 + 2 \sqrt{2}}{3 + 2\sqrt{2}}}$

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$\tt:\implies\: \: \: \: \: \: \: \: {\dfrac{(12 - 8\sqrt{2}) (3 + 2\sqrt{2})}{(3)^2 - (2\sqrt{2})^2}}$

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$\tt:\implies\: \: \: \: \: \: \: \: {\dfrac{36 + 24\sqrt{2} - 24\sqrt{2} - 32}{9 - 8}}$

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$\tt:\implies\: \: \: \: \: \: \: \: {\dfrac{36 - 32}{1}}$

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$\frak:\implies\: \: \: \: \: \: \: \: {\underline{\boxed{\purple{4}}}} \star$

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$\dag\small{\underline{\sf{Hence,\: the\: required\: value\: is\: 4.}}}$

Answer 2

Solution!!

x = 1 - √2

To find :- (x - 1/x)²

$\sf \left(1-\sqrt{2}-\dfrac{1}{1-\sqrt{2}}\right)^{2}$

Rationalise the denominator.

$\sf = \left(1-\sqrt{2}-\left(\dfrac{1}{1-\sqrt{2}}\times \dfrac{1+\sqrt{2}}{1+\sqrt{2}}\right)\right)^{2}$

$\sf = \left(1-\sqrt{2}-\left(\dfrac{1(1+\sqrt{2})}{(1-\sqrt{2})(1+\sqrt{2})}\right)\right)^{2}$

Use the identity → (a - b)(a + b) = a² - b².

$\sf = \left(1-\sqrt{2}-\left(\dfrac{1+\sqrt{2}}{(1)^{2}-(\sqrt{2})^{2}}\right)\right)^{2}$

$\sf = \left(1-\sqrt{2}-\left(\dfrac{1+\sqrt{2}}{1-2}\right)\right)^{2}$

$\sf = \left(1-\sqrt{2}-\left(\dfrac{1+\sqrt{2}}{-1}\right)\right)^{2}$

$\sf = \left(1-\sqrt{2}-\left(-\left(1+\sqrt{2}\right)\right)\right)^{2}$

Opening the brackets and changing the sign.

$\sf = \left(1-\sqrt{2}+\left(1+\sqrt{2}\right)\right)^{2}$

$\sf = \left(1-\sqrt{2}+1+\sqrt{2}\right)^{2}$

Grouping the like term.

$\sf = \left(1+1-\sqrt{2}+\sqrt{2}\right)^{2}$

$\sf = \left(1+1\right)^{2}$

$\sf = \left(2\right)^{2}$

$\sf = 4$