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Question
If `x + 1/x = sqrt5`, find the value of `x^2 + 1/x^2` and `x^4 + 1/x^4`
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Solution
We have
`(x + 1/x)^2 = x^2 + 1/x^2 + 2 xx x xx 1/x`
`=> (x + 1/x)^2 = x^2 + 1/x^2 + 2`
`=> (sqrt5)^2 = x^2 + 1/x^2 + 2` [∵ `x + 1/x = sqrt5`]
`=> 5 = x^2 + 1/x^2 + 2`
`=> x^2 + 1/x^2 = 3` ......(1)
Now `(x^2 + 1/x^2)^2 = x^4 + 1/x^4 + 2 xx x^2 xx 1/x^2`
`=> (x^2 + 1/x^2)^2 = x^4 + 1/x^4 + 2`
`=> 9 = x^2 + 1/x^4 + 2` [∵ `x^2 + 1/x^2 = 3`]
`=> x^4 + 1/x^4 = 7`
Hence `x^2 + 1/x^2 = 3; x^4 + 1/x^4 = 7`
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