Question

plz solve this question​

Answer 1

Answer:

given x =$\frac{\sqrt{5} -2}{\sqrt{5}+2 }$ ,y=$\frac{\sqrt{5} +2 }{\sqrt{5} -2}$

To find $x^{2} +y^{2} +xy$

 =  ( $\frac{\sqrt{5} -2}{\sqrt{5}+2 }$ ) ²+ ( $\frac{\sqrt{5} +2 }{\sqrt{5} -2}$ )² +$\frac{\sqrt{5} -2}{\sqrt{5}+2 }$ x $\frac{\sqrt{5} +2 }{\sqrt{5} -2}$

multiplying the conjugate

x  = ( $\frac{\sqrt{5} -2}{\sqrt{5}+2 }$ × $\frac{\sqrt{5} -2}{\sqrt{5} -2}$  )

    =  ((√ 5 -2)²)/(√5² -2²)

    =  (5 - 2×2×√5 + 4 )/(5-4)

    =(9 -4√5 ) /1

    =9 -4√5

x² = (9 -4√5)²

   = 81- 8√5 +16×5

    = 81  - 8√5 + 80

    = 161 -8√5

y =( $\frac{\sqrt{5} +2 }{\sqrt{5} -2}$ × $\frac{\sqrt{5} + 2}{\sqrt{5} +2}$ )

 = (√5 + 2)²  / (√5² - 2²)

 =( 5 + 2×2√5 +4) /(5-4)

 = 9+4√5

y² = (9+4√5)²

   = 81 + 8√5 + 16×5

   = 161 +8√5

x×y = 1 since numerator and denominator cancels each other

Therefore   x²  +y² + x×y = 161 -8√5 + 161 +8√5 + 1

                                      = 161 +161 +1

                                      = 323 (Answer)