If x = √3 - √2 /√3 + √2 and y = √3 + √2 / √3 - √2 , then x² + xy + y² =
A. 101
B. 99
C. 98
D. 102
Answers
Given : x = √3 - √2 /√3 + √2 and y = √3 + √2 / √3 - √2
To find : value of x² + xy + y²
Solution :
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
Hence the value of x² + xy + y² is 99.
Among the given options option (B) 99 is correct.
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Answer:
Step-by-step explanation:
To find : value of x² + xy + y²
Solution :
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