Question

if a=√3+√2 then prove that a^2+1/a^2=10
​

Answer 1

GIVEN :-

• a = √3 + √2
• a² + 1/a² = 10

TO PROVE :-

• a² + 1/a² = 10

SOLUTION :-

$\\ : \implies \tt \: a = \sqrt{3 } + \sqrt{2}$

$\red \bigstar \: \blue{ \tt \: Squaring \: both \: sides}$

$: \implies \tt (a) ^{2} = \bigg( \sqrt{3 } + \sqrt{2} \bigg) ^{2}$

$\red \bigstar \blue{ \tt \: using \: identity : - \: (a + b) ^{2} = a ^{2} +b ^{2} + 2ab }$

$\tt : \implies \: a ^{2} = \big( \sqrt{3} \big) ^{2} + \big( \sqrt{2} \big) ^{2} + 2 \times \sqrt{3} \times \sqrt{2}$

$: \implies \tt a ^{2} = 3 + 2 + 2 \sqrt{6}$

$: \implies \boxed{ \red{\tt a ^{2} =5 + 2 \sqrt{6} }}$

• Now let's calculate the value of a² + 1/a².

$\\ : \implies \tt a ^{2} + \dfrac{1}{a^{2} }$

$\red \bigstar \blue{ \tt \: \: put \: \: a ^{2} = 5 + 2 \sqrt{6} }$

$: \implies \tt 5 + 2 \sqrt{6} + \dfrac{1}{5 + 2 \sqrt{6} }$

$: \implies \tt \dfrac{ \bigg(5 + 2 \sqrt{6} \bigg) \bigg(5 + 2 \sqrt{6} \bigg) + 1 }{5 + 2 \sqrt{6} }$

$: \implies \tt \dfrac{ \bigg(5 + 2 \sqrt{6} \bigg) ^{2} + 1 }{5 + 2 \sqrt{6} }$

$\red \bigstar \blue{ \tt \: using \: identity : - \: (a + b) ^{2} = a ^{2} +b ^{2} + 2ab }$

$\tt \: : \implies \dfrac{ \bigg \{\big(5 \big) ^{2} + \big(2 \sqrt{6} \big) ^{2} + 2 \times 5 \times 2 \sqrt{6} \bigg \} + 1 }{5 + 2 \sqrt{6} }$

$: \implies \tt \: \dfrac{ \bigg(25 + 4 \times 6 + 20 \sqrt{6} \bigg) + 1 }{5 + 2 \sqrt{6} }$

$: \implies \tt \: \dfrac{ \bigg(25 + 24 + 20 \sqrt{6} \bigg) + 1}{5 + 2 \sqrt{6} }$

$: \implies \tt \: \dfrac{ 49 + 20 \sqrt{6} + 1 }{5 + 2 \sqrt{6} }$

$: \implies \tt \: \dfrac{50 + 20 \sqrt{6} }{5 + 2 \sqrt{6} }$

$\red \bigstar \blue{ \tt \:By \: \: Rationalising \: \: the \: \: denominator \: }$

$: \implies \tt \: \dfrac{ \bigg(50 + 20 \sqrt{6} \bigg) \bigg(5 - 2 \sqrt{6} \bigg)}{ \bigg(5 + 2 \sqrt{6} \bigg) \bigg(5 - 2 \sqrt{6} \bigg)}$

$\red \bigstar \blue{ \tt \: using \: identity : - \: (a + b) (a - b)= a ^{2} - b ^{2} }$

$\tt \: : \implies \dfrac{50 \bigg(5 - 2 \sqrt{6} \bigg) + 20 \sqrt{6} \bigg(5 - 2 \sqrt{6} \bigg)}{ \bigg(5 \bigg) ^{2} + \bigg(2 \sqrt{6} \bigg) ^{2} }$

$: \implies \tt \: \dfrac{250 - \cancel{ 100 \sqrt{6} +100 \sqrt{6}} - 240}{25 - 4 \times 6}$

$: \implies \tt \: \dfrac{250 - 240}{25 - 24}$

$: \implies \tt \: \dfrac{10}{1}$

$: \implies \tt \: \boxed{\tt a ^{2} + \dfrac{1}{a^{2} } = 10}$

HENCE PROVED ✅

Answer 2

$➔ \boxed{ \underline{ \tt \green{ GIVEN}}}$

$\tt↬ \blue{a = \sqrt{3} + \sqrt{2} }$

$➔ \boxed{ \underline{ \tt \red{ FIND}}}$

$\tt↬ \orange{ {a}^{2} + \frac{1}{ {a}^{2} } = 10 }$

$➔ \boxed{ \underline{ \tt \pink{ SOLUTION}}}$

$\tt⤞ \orange{ {a}^{2} + \frac{1}{ {a}^{2} } = 10 }......(i)$

$\tt⤞a = \sqrt{3} + \sqrt{2}$

$\tt \red{both \: side \: squaring}$

$\tt⤞ {a}^{2} = {( \sqrt{3} + \sqrt{2}) }^{2}$

$\tt by \: {(a + b)}^{2} = {a}^{2} + 2(a)(b) + {b}^{2} \: we \: have$

$\tt⤞ {a}^{2} = { \sqrt{3} }^{2} + 2( \sqrt{3} )( \sqrt{2} ) + { \sqrt{3} }^{2}$

$\tt⤞ {a}^{2} = 3 + 2 \sqrt{6} + 2$

$\tt⤞ {a}^{2} = 5 + 2 \sqrt{6}$

$\tt☄put \: value \: of \: {a}^{2} \: in \: eq(i)$

$\tt⤞ {a}^{2} + \frac{1}{ {a}^{2} } = 10$

$\tt⤞ 5 + 2 \sqrt{6} + \frac{1}{ 5 + 2 \sqrt{6} } = 10$

$\tt take \: L.C.M$

$\tt⤞ \frac{(5 + 2 \sqrt{6} )(5 + 2 \sqrt{6}) + 1 }{5 + 2 \sqrt{6} } = 10$

$\tt we \: can \: write(5 + 2 \sqrt{6} )(5 + 2 \sqrt{6} ) \: as \\ \tt {(5 + 2 \sqrt{6}) }^{2} \: so,$

$\tt⤞ \frac{(5 + 2 \sqrt{6} {)}^{2} + 1 }{5 + 2 \sqrt{6} } = 10$

$\tt by \: using{(a + b)}^{2} = {a}^{2} + 2(a)(b) + {b}^{2} \: we \: have$

$\tt⤞ \frac{ {5}^{2} + 2(5)(2 \sqrt{6} ) + {(2 \sqrt{6} )}^{2} + 1 }{5 + 2 \sqrt{6} } = 10$

$\tt⤞ \frac{ 25+ 20\sqrt{6} + 24 + 1 }{5 + 2 \sqrt{6} } = 10$

$\tt⤞ \frac{ 50+ 20\sqrt{6} }{5 + 2 \sqrt{6} } = 10$

$\tt now \: rationalising \: the \: denominator$

$\tt⤞ \frac{ 50+ 20\sqrt{6} }{5 + 2 \sqrt{6} } \times \frac{5 - 2 \sqrt{6} }{5 - 2 \sqrt{6} } = 10$

$\tt by \: using (a + b)(a - b) = {a}^{2} - {b}^{2} \\ \tt we \: have$

$\tt ⤞ \frac{50(5 - 2 \sqrt{6}) + 20 \sqrt{6}(5 - 2 \sqrt{6} ) }{ {5}^{2} - ({2 \sqrt{6}) }^{2} } = 10$

$\tt ⤞ \frac{250 - 100 \sqrt{6} + 100 \sqrt{6} - 240}{25 - 24} = 10$

$\tt ⤞ \frac{250 \cancel{- 100 \sqrt{6}} \cancel{+ 100 \sqrt{6} } - 240}{25 - 24} = 10$

$\tt ⤞ \frac{250 - 240}{25 - 24} = 10$

$\tt ⤞ \frac{10}{1} = 10$

$\tt ⤞ 10 = 10$

$\tt❂L.H.S=R.H.S$

$\tt⚝HENCE, \underline{\underline{PROVED}}$