Question

14.if a=√3+√2/√3-√2 and b=√3-√2/√3+√2 find the value of a2+b2​

Answer 1

Answer:-

Given:-

$\sf \: a = \dfrac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \: \: and \: \: b = \dfrac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{3} }$

We know that,

a² + b² = (a + b)² - 2ab

So,

$\implies \sf \: {a}^{2} + {b}^{2} = \bigg(\dfrac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } + \dfrac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \bigg) ^{2} - 2 \times \dfrac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \times \dfrac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \\ \\ \\ \implies \sf \: {a}^{2} + {b}^{2} = \bigg( \frac{( \sqrt{3} + \sqrt{2} ) ^{2} +(\sqrt{3} - \sqrt{2} ) ^{2} }{( \sqrt{3} - \sqrt{2} )( \sqrt{3} + \sqrt{2} ) } \bigg) ^{2} - 2(1)$

using (a + b)(a - b) = a² - b² ; (a + b)² = a² + b² + 2ab ; (a - b)² = a² + b² - 2ab we get,

$\implies \sf \: {a}^{2} + {b}^{2} = \bigg( \frac{{( \sqrt{3}) }^{2} + {( \sqrt{2}) }^{2} + 2( \sqrt{3})( \sqrt{2})+ {( \sqrt{3}) }^{2} + {( \sqrt{2}) }^{2} - 2( \sqrt{3})( \sqrt{2})}{ ( \sqrt{3} ) ^{2} - (\sqrt{2} ) ^{2} } \bigg) ^{2} - 2 \\ \\ \\ \implies \sf \: {a}^{2} + {b}^{2} = \bigg( \frac{3 + 2 + 2 \sqrt{6} + 3 + 2 - 2 \sqrt{6} }{ 3 - 2} \bigg) ^{2} - 2\\ \\ \\ \implies \sf \: {a}^{2} + {b}^{2} = (10) ^{2} - 2 \\ \\ \\\implies \sf \: {a}^{2} + {b}^{2} = 100 - 2 \\ \\ \\ \\\implies \boxed{ \sf \: {a}^{2} + {b}^{2} = 98}$

Answer 2

$\begin{gathered}\begin{gathered}\bf \: Given\:- \begin{cases} &\sf{a = \dfrac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } } \\ \\ &\sf{b = \dfrac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} }} \end{cases}\end{gathered}\end{gathered}$

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$\begin{gathered}\begin{gathered}\bf \: To \: find \:  - \begin{cases} &\sf{ {a}^{2} + {b}^{2} }  \end{cases}\end{gathered}\end{gathered}$

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$\begin{gathered}\Large{\bold{\green{\underline{Formula \: Used \::}}}} \end{gathered}$

$\rm :\implies\: \boxed{ \pink{ \bf \: {(x + y)}^{2} \: = \tt \: {x}^{2} + {y}^{2} + 2xy }}$

$\rm :\implies\: \boxed{ \pink{ \bf \: {(x - y)}^{2} \: = \tt \: {x}^{2} + {y}^{2} - 2xy }}$

$\rm :\implies\: \boxed{ \pink{ \bf \: {x}^{2} - {y}^{2} \: = \tt \: (x + y)(x - y)}}$

$\rm :\implies\: \boxed{ \pink{ \bf \: {(x + y)}^{2} + {(x - y)}^{2} \: = \tt \: 2( {x}^{2} + {y}^{2}) }}$

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

Consider,

$\rm :\implies\:a \: = \dfrac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} }$

On rationalizing the denominator, we get

$\rm :\implies\:a = \dfrac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \times \dfrac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} + \sqrt{2} }$

$\rm :\implies\:a = \dfrac{ {( \sqrt{3} + \sqrt{2} )}^{2} }{ {( \sqrt{3} )}^{2} - {( \sqrt{2}) }^{2} }$

$\rm :\implies\:a = \dfrac{3 + 2 + 2 \times \sqrt{3} \times \sqrt{2} }{3 - 2}$

$\rm :\implies\: \boxed{ \pink{ \bf \: a \: = \tt \: 5 + 2 \sqrt{6} }}$

Now,

Consider,

$\rm :\implies\:b \: = \: \dfrac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} }$

On rationalizing the denominator, we get

$\rm :\implies\:b = \dfrac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \times \dfrac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} - \sqrt{2} }$

$\rm :\implies\:b = \dfrac{ {( \sqrt{3} - \sqrt{2} ) }^{2} }{ {( \sqrt{3}) }^{2} - {( \sqrt{2} )}^{2} }$

$\rm :\implies\:b = \dfrac{3 + 2 - 2 \times \sqrt{3} \times \sqrt{2} }{3 - 2}$

$\rm :\implies\: \boxed{ \pink{ \bf \: b \: = \tt \: 5 - 2 \sqrt{6} }}$

Now,

Consider,

$\longrightarrow \green{ \bf \: {a}^{2} + {b}^{2} }$

On substituting the values of a and b, we get

$\rm :\implies\: {(5 + 2 \sqrt{6}) }^{2} + {(5 - 2 \sqrt{6}) }^{2}$

$\rm :\implies\:2 \{ {5}^{2} + {(2 \sqrt{6}) }^{2} \}$

$\rm :\implies\:2(25 + 24)$

$\rm :\implies\:2 \times 49$

$\rm :\implies\:98$

$\rm :\implies\: \boxed{ \pink{ \bf \: {a}^{2} + {b}^{2} \: = \tt \: 98 }}$