[tex]\large \red{\bf{ Ques \pink{tion}}}[/tex]
[tex] \bf If\begin{cases} \sf x = \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \\ \sf \: and \\ \sf y = \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \end{cases}[/tex]
Then Find the value of-
[tex]\large \purple{\bf{ {x}^{2} + {y}^{2} + xy }}[/tex]
Answers & Comments
Answer:
=x²-y²-x-y
=(x²-y²)-1(x+y)
=(x+y)(x-y)-1(x+y). ,(x²-y²)=(x+y)(x-y)
=(x+y){(x-y)-1}
=(x+y)(x-y+1)
I hope it helps you,BDW my name is Souvik Das
Verified answer
[tex]\large\underline{\sf{Solution-}}[/tex]
Given that
[tex]\sf \: x = \dfrac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \\ \\ [/tex]
On rationalizing the denominator, we get
[tex]\sf \: x = \dfrac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \times \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} - \sqrt{2}} \\ \\ [/tex]
[tex]\sf \: x= \dfrac{ {(\sqrt{3} - \sqrt{2})}^{2} }{ {( \sqrt{3} )}^{2} - {( \sqrt{2} )}^{2} } \\ \\ [/tex]
[tex]\sf \: x = \dfrac{ {( \sqrt{3} )}^{2} + {( \sqrt{2} )}^{2} - 2 \times \sqrt{3} \times \sqrt{2} }{3 - 2} \\ \\ [/tex]
[tex]\sf \: x = 3 + 2 - 2 \sqrt{6} \\ \\ [/tex]
[tex]\bf\implies \:x = 5 - 2 \sqrt{6} - - - (1) \\ \\ [/tex]
Further, given that
[tex]\sf \: y = \dfrac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \\ \\ [/tex]
On rationalizing the denominator, we get
[tex]\sf \: y = \dfrac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \times \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} + \sqrt{2}} \\ \\ [/tex]
[tex]\sf \: y= \dfrac{ {(\sqrt{3} + \sqrt{2})}^{2} }{ {( \sqrt{3} )}^{2} - {( \sqrt{2} )}^{2} } \\ \\ [/tex]
[tex]\sf \: y = \dfrac{ {( \sqrt{3} )}^{2} + {( \sqrt{2} )}^{2} + 2 \times \sqrt{3} \times \sqrt{2} }{3 - 2} \\ \\ [/tex]
[tex]\sf \: y = 3 + 2 + 5 \sqrt{6} \\ \\ [/tex]
[tex]\bf\implies \:y= 5 + 2 \sqrt{6} - - - (2) \\ \\ [/tex]
Now, Consider
[tex]\sf \: {x}^{2} + {y}^{2} + xy \\ \\ [/tex]
can be rewritten as
[tex]\sf \: = \: {x}^{2} + {y}^{2} + 2xy - xy \\ \\ [/tex]
[tex]\sf \: = \: {(x + y)}^{2} - xy \\ \\ [/tex]
[tex]\sf \: = \: {(5 - 2 \sqrt{6} + 5 + 2 \sqrt{6} )}^{2} - \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \times \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}} \\ \\ [/tex]
[tex]\sf \: = \: {(10)}^{2} - 1 \\ \\ [/tex]
[tex]\sf \: = \: 100 - 1 \\ \\ [/tex]
[tex]\sf \: = \: 99 \\ \\ [/tex]
Hence,
[tex]\rm\implies \:\boxed{ \sf{ \: {x}^{2} + {y}^{2} + xy = 99 \: }} \\ \\ [/tex]
[tex]\rule{190pt}{2pt}[/tex]
Basic Identities Used :-
[tex]\boxed{ \sf{ \: {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy \: \: }} \\ \\ [/tex]
[tex]\boxed{ \sf{ \: {(x - y)}^{2} = {x}^{2} + {y}^{2} - 2xy \: \: }} \\ \\ [/tex]
[tex]\boxed{ \sf{ \:(x + y)(x - y) = {x}^{2} - {y}^{2} \: }} \\ \\ [/tex]