Question

please solve this question ​

Answer 1

Given :

• $\sf a = \dfrac{\sqrt{5}+\sqrt{2}}{\sqrt{5}-\sqrt{2}}$
• $\sf b = \dfrac{\sqrt{5} -\sqrt{2}}{\sqrt{5}+\sqrt{2}}$

To prove :

• $\sf 3a^2 +4ab -3b^2 = 4 + \dfrac{56}{3}\sqrt{10}$

Solution :

Firstly we will rationalise a and b to solve the question further.

$\sf \implies a = \dfrac{\sqrt{5}+\sqrt{2}}{\sqrt{5}-\sqrt{2}}$

$\sf \implies a = \dfrac{\sqrt{5}+\sqrt{2}}{\sqrt{5}-\sqrt{2}} \times \dfrac{ \sqrt{5} + \sqrt{2} }{ \sqrt{5} + \sqrt{2} }$

$\sf \implies a = \dfrac{(\sqrt{5}+\sqrt{2})^{2} }{(\sqrt{5})^{2} -(\sqrt{2})^{2} }$

$\sf \implies a = \dfrac{5 + 2 + 2 \sqrt{10} }{5 - 2 }$

$\boxed{ \sf \implies a = \dfrac{7 + 2 \sqrt{10} }{3}}$

Similarly, rationalising b

$\sf \implies b= \dfrac{\sqrt{5} - \sqrt{2}}{\sqrt{5} + \sqrt{2}}$

$\sf \implies b= \dfrac{\sqrt{5} - \sqrt{2}}{\sqrt{5} + \sqrt{2}} \times \dfrac{ \sqrt{5} - \sqrt{2} }{ \sqrt{5} - \sqrt{2} }$

$\sf \implies b= \dfrac{(\sqrt{5} - \sqrt{2})^2}{(\sqrt{5})^{2} - (\sqrt{2})^{2} }$

$\sf \implies b= \dfrac{ {( \sqrt{5}) }^{2} + (\sqrt{2} )^{2} - 2( \sqrt{5} )( \sqrt{2}) }{5 - 2 }$

$\sf \implies b= \dfrac{ 5 + 2- 2\sqrt{10} }{3}$

$\boxed{ \sf \implies b= \dfrac{ 7- 2\sqrt{10} }{3} }$

We have to prove that,

$\sf\implies 3a^2 +4ab -3b^2 = 4 + \dfrac{56}{3}\sqrt{10}$

Consider LHS and substitute values of a and b

$\sf \implies 3a^2 +4ab -3b^2$

$\sf \implies 3\left[ \dfrac{7 + 2 \sqrt{10} }{3} \right]^2 +4\left[ \dfrac{7 + 2 \sqrt{10} }{3} \right]\left[ \dfrac{7 - 2 \sqrt{10} }{3} \right]-3\left[ \dfrac{7 + 2 \sqrt{10} }{3} \right]^{2}$

$\sf \implies 3 \times \dfrac{(7 + 2 \sqrt{10})^{2} }{9} +4\left[ \dfrac{(7 + 2 \sqrt{10})(7 - 2 \sqrt{10} ) }{9} \right]-3 \times \dfrac{(7 + 2 \sqrt{10})^{2} }{9}$

$\sf \implies \dfrac{( {7)}^{2} + {(2 \sqrt{10} )}^{2} + 2(7)(2 \sqrt{10} )}{3} +4\left[ \dfrac{(7)^{2} -( 2 \sqrt{10})^{2} }{9} \right]- \dfrac{(7)^{2} + (2 \sqrt{10})^{2} - 2(7)(2 \sqrt{10}) }{3}$

$\sf \implies \dfrac{49 + 40 + 28 \sqrt{10}}{3} + 4 \times \left[ \dfrac{49 - 40}{9} \right]- \dfrac{49 + 40 - 28 \sqrt{10}}{3}$

$\sf \implies \dfrac{89+ 28 \sqrt{10}}{3} + 4 \times \left[ \dfrac{9}{9} \right]- \dfrac{89- 28 \sqrt{10}}{3}$

$\sf \implies \dfrac{89+ 28 \sqrt{10}}{3} + 4 - \dfrac{89- 28 \sqrt{10}}{3}$

$\sf \implies \dfrac{89+ 28 \sqrt{10} - (89 - 28 \sqrt{10}) }{3} + 4$

$\sf \implies \dfrac{89+ 28 \sqrt{10} - 89 + 28 \sqrt{10} }{3} + 4$

$\sf \implies \dfrac{56\sqrt{10} }{3} + 4$

Hence the required result is proved.

Answer 2

$\color{red} \underline { \underline{ \bf {STEP-BY-STEP \: \: EXPLANATION :}}}$

$\implies\tt a = \frac{ \sqrt{5} + \sqrt{2} }{ \sqrt{5} - \sqrt{2} } \: \: \: ...(given)$

$\implies\tt a = \frac{ \sqrt{5} + \sqrt{2} }{ \sqrt{5} - \sqrt{2} } \times \frac{ \sqrt{5} + \sqrt{2} }{ \sqrt{5} + \sqrt{2} }$

$\bf [ (a + b)(a - b) = {a}^{2} - {b}^{2} ]$

$\implies\tt a = \frac{( \sqrt{5} + \sqrt{2} ) ^{2} }{( \sqrt{5} - \sqrt{2}) ( \sqrt{5} + \sqrt{2}) }$

$\bf [(a + b) ^{2} = {a}^{2} + 2ab + {b}^{2} ]$

$\implies \tt a = \frac{(\sqrt{5})^{2} + 2 \sqrt{5} . \sqrt{2} + (\sqrt{2}) ^{2} }{ { (\sqrt{5} )}^{2} - {( \sqrt{2} )}^{2} }$

$\implies \tt a = \frac{5 + 2 \sqrt{10} + 2 }{5 - 2}$

$\implies \boxed {\tt a = \frac{7+ 2 \sqrt{10}}{3} }$

$\implies \tt b= \frac{ \sqrt{5} - \sqrt{2} }{ \sqrt{5} + \sqrt{2} } \: \: \: ...(given)$

$\implies \tt b = \frac{ \sqrt{5} - \sqrt{2} }{ \sqrt{5} + \sqrt{2} } \times \frac{ \sqrt{5} - \sqrt{2} }{ \sqrt{5} - \sqrt{2} }$

$\implies \tt b= \frac{( \sqrt{5} - \sqrt{2} ) ^{2} }{( \sqrt{5} + \sqrt{2}) ( \sqrt{5} - \sqrt{2}) }$

$\bf [(a - b) ^{2} = {a}^{2} - 2ab + {b}^{2} ]$

$\implies \tt b= \frac{(\sqrt{5})^{2} - 2 \sqrt{5} . \sqrt{2} + (\sqrt{2}) ^{2} }{ { (\sqrt{5} )}^{2} - {( \sqrt{2} )}^{2} }$

$\implies \boxed{ \tt b = \frac{7 - 2 \sqrt{10}}{3} }$

$\implies \tt 3 {a}^{2} + 4ab - 3 {b}^{2} = 4 + \frac{56}{3} \sqrt{10}$

$\implies \tt 3 {( \frac{7 + 2 \sqrt{10} }{3}) }^{2} + 4( \frac{7 + 2 \sqrt{10} }{3}) ( \frac{7 - 2 \sqrt{10} }{3}) - 3 {( \frac{7 - 2 \sqrt{10} }{3} )}^{2} = 4 + \frac{56}{3} \sqrt{10}$

$\implies \tt 3 {( \frac{7 + 2 \sqrt{10} }{3}) }( \frac{7 + 2 \sqrt{10} }{3}) + \frac{4(7 + 2 \sqrt{10} )(7 - 2 \sqrt{10}) }{9} - 3 {( \frac{7 - 2 \sqrt{10} }{3} )}( \frac{7 - 2 \sqrt{10} }{3} ) = 4 + \frac{56}{3} \sqrt{10}$

$\implies \tt \frac{ {7}^{2} + (2 \sqrt{10})^{2} + 2(7)(2 \sqrt{10} ) }{3} + \frac{4( {7}^{2} - {(2 \sqrt{10}) }^{2} )}{9} - \frac{ {7}^{2} + (2 \sqrt{10})^{2} - 2(7)(2 \sqrt{10} ) }{3} = 4 + \frac{56}{3} \sqrt{10}$

$\implies \tt \frac{ 49 + 40 + 28\sqrt{10} }{3} + \frac{4( 49 - 40 )}{9} - \frac{49 +40 - 28 \sqrt{10} }{3} = 4 + \frac{56}{3} \sqrt{10}$

$\implies \tt \frac{ 89+ 28\sqrt{10} }{3} + \frac{4( 9)}{9} - \frac{89 - 28 \sqrt{10} }{3} = 4 + \frac{56}{3} \sqrt{10}$

$\implies \tt 4 + \frac{ 89+ 28\sqrt{10} }{3} - \frac{89 - 28 \sqrt{10} }{3} = 4 + \frac{56}{3} \sqrt{10}$

$\implies \tt 4 + \frac{ 89+ 28\sqrt{10} - (89 - 28 \sqrt{10} ) }{3} = 4 + \frac{56}{3} \sqrt{10}$

$\implies \tt 4 + \frac{ 89+ 28\sqrt{10} - 89 + 28 \sqrt{10} }{3} = 4 + \frac{56}{3} \sqrt{10}$

$\implies \tt 4 + \frac{ 56\sqrt{10} }{3} = 4 + \frac{56}{3} \sqrt{10}$

$\implies \bf LHS = RHS$

$\bf \underline{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:Hence \: Proved }$