Question

if x= 7+4 root 3 and xy= 1 find 1/x^2+1/y^2

Answer 1

Required Answer:-

Given:

• x = 7 + 4√3
• xy = 1

To Find:

• The value of 1/x² + 1/y²

Solution:

Given that,

→ x = 7 + 4√3

As xy = 1,

→ y = 1/x

→ y² = 1/x²

→ 1/y² = x²

So,

→ 1/x² + 1/y² is same as x² + 1/x²

Calculating the value of 1/x,..

→ 1/x = 1/(7 + 4√3)

→ 1/x = 1 × (7 - 4√3)/[(7 + 4√3)(7 - 4√3)]

→ 1/x = (7 - 4√3)/(7² - (4√3)²)

→ 1/x = (7 - 4√3)/(49 - 48)

→ 1/x = 7 - 4√3

So, the value of 1/x² + 1/y² will be,

= 1/x² + x² (As x = 1/y)

= x² + 1/x²

= (7 + 4√3)² + (7 - 4√3)²

Applying identity (a + b)² = a² + 2ab + b² and (a - b)² = a² - 2ab + b², we get,

= (7)² + 2 × 7 × 4√3 + (4√3)² + (7)² - 2 × 7 × 4√3 + (4√3)²

= (7)² + (7)² + (4√3)² + (4√3)²

= 49 + 49 + 48 + 48

= 98 + 96

= 194

→ 1/x² + 1/y² = 194

Answer:

• 1/x² + 1/y² = 194

•••♪