Question

If x=√3-√2/√3+√2and y=√3+√2/√3-√2 find the value of x2+y2+xy​

Answer 1

Gɪᴠᴇɴ :-

$\bf\:x=\dfrac{\sqrt{3}-\sqrt{2}}{ \sqrt{3}+\sqrt{2} }$

$\bf\:y=\dfrac{\sqrt{3} + \sqrt{2}}{ \sqrt{3} - \sqrt{2} }$

Tᴏ Fɪɴᴅ :-

• x² + y² + xy = ?

Sᴏʟᴜᴛɪᴏɴ :-

Rationalizing x & y we get :-

$\purple\longmapsto\tt\:x=\dfrac{\sqrt{3}-\sqrt{2}}{ \sqrt{3}+\sqrt{2}} \times \: \dfrac{\sqrt{3}-\sqrt{2}}{ \sqrt{3} - \sqrt{2}} \\ \\ \purple\longmapsto\tt\:x= \dfrac{(\sqrt{3}-\sqrt{2})^{2}}{( \sqrt{3})^{2} - (\sqrt{2})^{2}} \\ \\\purple\longmapsto\tt\:x= \dfrac{3 + 2 - 2\sqrt{6}}{3 - 2} \\ \\\purple\longmapsto\boxed{\tt\:x=5 - 2 \sqrt{6}}$

Similarly,

$\purple\longmapsto\tt\:y=\dfrac{\sqrt{3} + \sqrt{2}}{ \sqrt{3} - \sqrt{2}} \times \: \dfrac{\sqrt{3} +\sqrt{2}}{ \sqrt{3}+ \sqrt{2}} \\ \\ \purple\longmapsto\tt\:y= \dfrac{(\sqrt{3} + \sqrt{2})^{2}}{( \sqrt{3})^{2} - (\sqrt{2})^{2}} \\ \\\purple\longmapsto\tt\:y= \dfrac{3 + 2 +2\sqrt{6}}{3 - 2} \\ \\\purple\longmapsto\boxed{\tt\:y=5+2 \sqrt{6}}$

And,

$\purple\longmapsto\tt\:x \times y=\dfrac{\sqrt{3} -\sqrt{2}}{ \sqrt{3}+\sqrt{2}} \times \: \dfrac{\sqrt{3} + \sqrt{2}}{ \sqrt{3} - \sqrt{2}} \\ \\ \purple\longmapsto\boxed{\tt\:xy=1}$

Putting All Values Now, we get :-

$\red\longrightarrow\bf\:{x}^{2}+{y}^{2} + xy \\ \\\red\longrightarrow\bf\:(x + y)^{2} - xy \\ \\ \red\longrightarrow\bf\:(5 - 2 \sqrt{6} + 5 + 2 \sqrt{6})^{2} - 1 \\ \\\red\longrightarrow\bf\:(10)^{2} - 1 \\ \\\red\longrightarrow\bf\:100 - 1 \\ \\\red\longrightarrow\red{\large\boxed{\bf\:99}}$

$\rule{200}{4}$

$\bold{\boxed{\huge{\boxed{\orange{\small{\boxed{\huge{\red{\bold{\huge\bold{\red{\ddot{\smile}}}}}}}}}}}}}$

Answer 2

$\dag\ \large\underline{\mathbb{GIVEN :-}}$

$\bullet\ \sf{x=\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}}$

$\bullet\ \sf{y=\dfrac{\sqrt{3}+\sqrt{2} }{\sqrt{3}-\sqrt{2} } }$

$\dag\ \large\underline{\mathbb{TO\ FIND :-}}$

$\sf{x^2+y^2+xy}$

$\dag\ \large\underline{\mathbb{CONCEPT\ USED :-}}$

$\sf{x\ and\ y\ are\ rationalised\ to\ remove\ the\ roots}\\\sf{from\ the\ denominator.}$

$\dag\ \large\underline{\mathbb{SOLUTION :-}}$

$\underline{\sf{Rationalising\ x }}$

$\sf{x=\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}}\\\\\\\sf{\implies x=\dfrac{(\sqrt{3}-\sqrt{2})(\sqrt{3}-\sqrt{2} )}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2} ) } }\\\\\\\sf{\implies x=\dfrac{(\sqrt{3}-\sqrt{2} )^2}{(\sqrt{3} )^2-(\sqrt{2} )^2} }\\\\\\\sf{\implies x=\dfrac{(\sqrt{3} )^2+(\sqrt{2} )^2-2\times \sqrt{3}\times \sqrt{2} }{3-2} }\\\\\\\sf{\implies x=3+2-2\sqrt{6} }\\\\\boxed{\sf{\implies x=5-2\sqrt{6} }}$

$\rule{200}2$

$\underline{\sf{Rationalising\ y }}$

$\sf{y=\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}}\\\\\\\sf{\implies y=\dfrac{(\sqrt{3}+\sqrt{2})(\sqrt{3}+\sqrt{2} )}{(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2} ) } }\\\\\\\sf{\implies y=\dfrac{(\sqrt{3}+\sqrt{2} )^2}{(\sqrt{3} )^2-(\sqrt{2} )^2} }\\\\\\\sf{\implies y=\dfrac{(\sqrt{3} )^2+(\sqrt{2} )^2+2\times \sqrt{3}\times \sqrt{2} }{3-2} }\\\\\\\sf{\implies y=3+2+2\sqrt{6} }\\\\\boxed{\sf{\implies y=5+2\sqrt{6} }}$

$\rule{200}2$

$\sf{x^2=(5-2\sqrt{6})^2 }\\\\\sf{\implies x^2=(5)^2+(2\sqrt{6})^2-2\times5\times2\sqrt{6}}\\\\\sf{\implies x^2=25+24-20\sqrt{6} }\\\\\sf{\implies x^2=49-20\sqrt{6} }$

$\rule{150}{1.5}$

$\sf{y^2=(5+2\sqrt{6})^2 }\\\\\sf{\implies y^2=(5)^2+(2\sqrt{6})^2+2\times5\times2\sqrt{6}}\\\\\sf{\implies x^2=25+24+20\sqrt{6} }\\\\\sf{\implies y^2=49+20\sqrt{6} }$

$\rule{150}{1.5}$

$\sf{xy=(5-2\sqrt{6})(5+2\sqrt{6}) }\\\\\sf{\implies xy=(5)^2-(2\sqrt{6} )^2}\\\\\sf{\implies xy=25-24}\\\\\sf{\implies xy=1}$

$\underline{\sf{Now\ putting\ all\ the\ values:-}}$

$\sf{x^2+y^2+xy=49-20\sqrt{6}+49+20\sqrt{6}+1}\\\\\sf{\longrightarrow x^2+y^2+xy=49+49+1}\\\\\boxed{\sf{\longrightarrow x^2+y^2+xy=99}}\:\:\:\:\:\: \mathbf{\cdots ANSWER}$