Question

If \[x = 7 + 4\sqrt{3}\] and xy =1, then \[\frac{1}{x^2} + \frac{1}{y^2} =\]

Options

• 64

• 134

• 194

• 1/49

Answer 1

Given that `x=7+4sqrt3`, `xy = 1`

Hence  y is given as 

 `y=1/x`

 `1/x = 1/(7+4sqrt3)`

We need to find  `1/x^2+ 1/y^2`

We know that rationalization factor for   `7+4sqrt3` is `7-4sqrt3`. We will multiply numerator and denominator of the given expression `1/(7+4sqrt3)`by,`7-4sqrt3` to get

`1/x = 1/(7+4sqrt3) xx (7-4sqrt3)/(7-4sqrt3)`

`= (7-4sqrt3)/((7)^2 (4sqrt3)^2) `

` = (7-4sqrt3)/(49 - 48)`

` = 7-4sqrt3`

Since `xy=1`so we have 

 `x=1/y`

Therefore,

`1/x^2 + 1/y^2 = ( 7 - sqrt3)^2 + (7+4sqrt3)^2`

` = 7^2 + (4sqrt3)^2 - 2 xx 7 xx 4sqrt3 + 7 ^2 +(4 sqrt3)^2 + 2 xx 7 xx 4sqrt3`

`= 49 + 48 - 14 sqrt3 + 49 +48 +14sqrt3`

`= 194`