## If x =√3-√2/√3+√2 y=√3+√2/√3-√2 so what is the value of x²+y²+xy

Question

If x =√3-√2/√3+√2 y=√3+√2/√3-√2 so what is the value of x²+y²+xy

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5 months 2022-01-06T17:51:38+00:00 2 Answers 0 views 0

We have, x = (√3 – √2 )/(√3 + √2) and y = (√3 + √2) / (√3 – √2)

On rationalising the denominator of x :

X = (√3 – √2 ) × (√3 – √2) /(√3 + √2) (√3 – √2)

x = (√3 – √2)² / (√3 + √2) (√3 – √2)

By Using Identity : (a – b)² = a² + b² – 2ab & (a + b)(a – b) = a² – b²

x = {√3² + √2² – 2 × √3 × √2}/ √3² – √2²

x = {3 + 2 – 2√6}/( 3 – 2)

x = 5 – 2√6 ………..(1)

Now ,

x² = (5 – 2√6)²

By Using Identity : (a – b)² = a² + b² – 2ab

x² = 5² + (2√6)² – 2 × 5 × 2√6

x² = 25 + 24 – 20√6

x² = 49 – 20√6 …………(2)

On rationalising the denominator of y :

y = (√3 + √2) × (√3 + √2)/ (√3 – √2) × (√3 + √2)

y = (√3 + √2)²/(√3 – √2) × (√3 + √2)

By Using Identity : (a + b)² = a² + b² + 2ab & (a + b)(a – b) = a² – b²

y = {√3² + √2² + 2 × √3 × √2}/ √3² – √2²

y = {3 + 2 + 2√6}/( 3 – 2)

y = 5 + 2√6 …………(3)

Now ,

y² = (5 + 2√6)²

By Using Identity : (a + b)² = a² + b² + 2ab

y² = 5² + (2√6)² + 2 × 5 × 2√6

y² = 25 + 24 + 20√6

y² = 49 + 20√6 …………(4)

Now,

From eq 1 & 3 :

xy = (5 – 2√6) ( 5 + 2√6)

By Using Identity : (a + b)(a – b) = a² – b²

xy = 5² – (2√6)²

xy = 25 – 24

xy = 1 …………(5)

Therefore , x² + xy + y²

From eq 2, 4 & 5 :

= 49 – 20√6 + 1 + 49 + 20√6

= 49 + 49 + 1

= 99

x² + xy + y² = 99

Step-by-step explanation:

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