Question

Real Numbers Class IX​

Answer 1

$\begin{gathered}\begin{gathered}\bf \: Given - \begin{cases} &\sf{a \: = \sqrt{2} + 1 } \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf \: To\:find\: - \begin{cases} &\sf{ \left( {(a \: + \: \dfrac{1}{a} } \right)}^{2}\\ &\sf{\left( {(a \: - \: \dfrac{1}{a} } \right)}^{2}  \end{cases}\end{gathered}\end{gathered}$

$\large\underline\purple{\bold{Solution :- }}$

Given that

$\rm :\implies\:a \: = \: \sqrt{2} + 1$

Now,

Consider,

$\rm :\implies\:\dfrac{1}{a}$

$\rm :\implies\:\dfrac{1}{ \sqrt{2} + 1}$

On rationalizing the denominator, we get

$\rm :\implies\:\dfrac{1}{ \sqrt{2} + 1} \times \dfrac{ \sqrt{2} - 1}{ \sqrt{2} - 1}$

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

$\rm :\implies\:\dfrac{ \sqrt{2} - 1}{2 - 1}$

$\rm :\implies\:\dfrac{ \sqrt{2} - 1 }{1}$

$\rm :\implies\: \boxed{ \pink{ \bf \:\dfrac{1}{a} \: = \tt \: \sqrt{2} - 1 }}$

Now,

Consider,

$\rm :\implies\: \red{ \bf \: \left( {(a \: + \: \dfrac{1}{a} } \right)}^{2}$

$\rm :\implies\: {( \sqrt{2} - 1 + \sqrt{2} + 1) }^{2}$

$\rm :\implies\: {(2 \sqrt{2} )}^{2}$

$\rm :\implies\:8$

So,

$\rm :\implies\: \boxed{ \pink{ \bf \: { \bigg(a \: + \: \dfrac{1}{a} \bigg) }^{2} \: = \tt \:8 }}$

Consider,

$\green{\rm :\implies\:{ \bigg(a \: - \: \dfrac{1}{a} \bigg) }^{2}}$

$\rm :\implies\: {( \sqrt{2} + 1 - \sqrt{2} + 1) }^{2}$

$\rm :\implies\: {(2)}^{2}$

$\rm :\implies\:4$

$\rm :\implies\: \boxed{ \pink{ \bf \:{ \bigg(a \: - \: \dfrac{1}{a} \bigg) }^{2} \: = \tt \:4 }}$

Answer 2

$\bf \: Given$

$\to \tt \: a = \sqrt{2} + 1$

$\bf \: To \: Find$

$\tt \to \bigg(a + \dfrac{1}{ {a}^{} } \bigg) ^{2} \\ \tt \to \bigg(a - \dfrac{1}{ {a}^{} } \bigg) ^{2}$

$\bf \: Now \: Take$

$\tt \to \: a = \sqrt{2} + 1$

$\tt \to \: \dfrac{1}{a} = \dfrac{1}{ \sqrt{2} + 1}$

$\tt \to \: \dfrac{1}{ \sqrt{2} + 1 } \times \dfrac{ \sqrt{2} - 1}{ \sqrt{2} - 1 }$

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

$\tt \to \: \dfrac{ \sqrt{2} - 1}{2 - 1}$

$\tt \to \: \sqrt{2} - 1 = \dfrac{1}{a}$

$\bf \:Now \: Find$

$\tt \to \bigg(a + \dfrac{1}{ {a}^{} } \bigg) ^{2} \\ \tt \to \bigg(a - \dfrac{1}{ {a}^{} } \bigg) ^{2}$

$\bf \: where$

$\tt \to \: a = \sqrt{2} + 1$

$\tt \to \: \dfrac{1}{a} = \sqrt{2} - 1$

$\tt \to \bigg(a + \dfrac{1}{ {a}^{} } \bigg) ^{2}$

$\tt \to( \sqrt{2} + 1 + \sqrt{2} - 1) {}^{2}$

$\tt \to(2 \sqrt{2} ) {}^{2} = 4 \times 2 = 8$

$\tt \to \bigg(a - \dfrac{1}{ {a}^{} } \bigg) ^{2}$

$\tt \to \:( \sqrt{2} + 1 - \sqrt{2} + 1) {}^{2}$

$\tt \to \: (2) {}^{2} = 4$