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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
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Solution
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`
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